Questions on Logic: Propositions, Conjunction, Disjunction, Implication answered by real tutors!

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I haven't taken formal logic, but it is possible that crops will be good even if it doesn't rain. This is like saying "If a quadrilateral is a square, it is a rectangle." However, the obverse "If a quadrilateral is a rectangle, it is a square" is not always true. It is true if and only if the relation is one-to-one.

However, the contrapositive is always true and equivalent to the original statement: "If crops are not good, it did not rain."

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in your problem, you have the following:

Let p, q, and r represent the following statements.
p: Jamie is on the train
q: Sylvia is at the park
r: Nigel is in the car
Construct a truth table for the following
a. ~q V p
b. (~p V q)↔ q

the variables involved are p and q and r
this means your truth table will have 2^3 = 8 rows.
you'll have 1 column for p and 1 column for q and 1 column for r
you should also construct a column for ~q and ~p since these will be involved in the statements.
if p is true, then ~p is false.
if q is true, then ~q is false.
if p is false, then ~p is true.
if q is false, then ~q is true.
make a column for (~q v p)
~q v p is true if either q is false or p is true.
it is only false the statement q is false is false and if the statement p is true is also false.
if the statement q is false is false, this means that q is true.
if the statement p is true is false, this means that p is false.
this means that ~q v p is false if q is true and p is false.
to understand this, you have to understand the truth table for A v B
that truth table looks like this:

               A           B        A v B
               T           T          T
               T           F          T
               F           T          T
               F           F          F

A v B is true in all cases except when both A and B are false.
now if you let ~q be equal to A and you let p be equal to B, then the truth table becomes:

               ~q          p        ~q v p
                T          T           T
                T          F           T
                F          T           T
                F          F           F

same truth table with same logic only the names of the variables have been changed which is totally legitimate.

you can see that ~q v p is only false when ~q is false and p is false.
you also know that ~q is false if q is true, so the equivalent statement becomes:
~q v p is false if q is true and p is false.

you construct your ~q v p column based on the OR rules as expressed above.
the ~q v p column is true if either ~q is true or if p is true or if both are true. if both are false, then the ~q v p column is false. you'll see this in the table.

you now want to construct another column for (~p v q).
this is another OR construction, only this time the statement is true if either ~p is true or q is true of both are true. the statement is false if ~p is false and q is false at the same time.

you now want to construct another column for (~p v q) <-> q
you will be comparing columns of (~p v q) and q in order to determine the validity of the statement (~p v q) <-> q

the logic for the if and only if statement is as follows:

                A            B        A <-> B
                T            T           T
                T            F           F
                F            T           F
                F            F           T

If A and B are both true, then the statement A <-> B is true.
if A and B are both false, then the statement A <-> B is true.
In other words, if they are both the same, the statement A <-> B is true.
if they are both different, then the statement A <-> B is false.
this includes:
A is true and B is false.
A is false and B is true.

if you let A = (~p v q) and if you let B = q, then this logic applies to the statement (~p v q) <-> q

the truth table for that would be as follows:

            (~p v q)         q        (~p v q) <-> q
                T            T                  T
                T            F                  F
                F            T                  F
                F            F                  T

when you construct your column for, you would follow this logic in setting up the column by comparing the columns for (~p v q) and q.

your final truth table is shown below:


here's a reference on truth tables if you're interested.


http://www.algebra.com/algebra/homework/Conjunction/THEO-2011-08-19.lesson

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Hello my questions is with this equation: ((x->y)^(y->z))->(x->z)
I'm suppose to prove that it is a tautology through a truth table. I was just wondering when you throw that third letter in there (z) how does the truth table start. Is it that there are now 6 rows instead of a normal 4 with two variables. like x would start with three T's and three F's? Thanks in advance
-------
Use the following pattern:
x.....y....z
t.....t....t
t.....t....f
t.....f....t
t.....f....f
f.....t....t
f.....t....f
f.....f....t
f.....f....f
========================
Cheers,
Stan H.

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divide the measure of each item in the recipe by 5
==================
Cheers,
Stan H.

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the law of transitivity means this:
if a implies b and b implies c then a implies c.
the statement above boils down to:
if i tell you the time, then my life will be miserable.
this assumes the chain of logic follows through to the end.
the chain is as follows:
if i tell you the time then we start chatting (a implies b)
if we start chatting, then we'll meet at the local pub (b implies c)
if we meet at the local pub then we'll discuss my family (c implies d)
if we discuss my family then you'll find out that my daughter is available (d implies e)
if you find out that my daughter is available then you'll want to marry her (e implies f)
if you want to marry her then my life will be miserable. (f implies g).
the law is:
if a implies b and b implies c, then a implies c.
we now have a implies c
the law then states:
if a implies c and c implies d, then a implies d.
we now have a implies d.
the law then states:
if a implies d and d implies e, than a implies e.
we now have a implies e.
the law then states:
if a implies e and e implies f, then a implies f.
we now have a implies f.
the law then states:
if a implies f and f implies g, then a implies g.
we now have a implies g.
this means that:
if i tell you the time then i will be miserable by the law of transitivity.

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Since there are no V's in the hat to begin with, the probability of not selecting a V is 100%.

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The inverse of the truth functional statement 'If P, then Q' is 'If not P, then not Q.' So the inverse of the given statement should be 'If I wash my clothes, then I will have something clean to wear.'

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~p -> (q V ~r)
                put T for p and q, F for r 
~T -> (T V ~F)
                change ~T to F and ~F to T
 F -> (T V T)
                change (T V T) to T  (V is T except when F on both sides) 
 F -> T
                 (-> is T except when T on the left and F on the right) 
    T

Edwin

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Write the statement in symbols using the p and q given below. Then construct a truth table for the symbolic statement and select the best match.
p = I eat too much. q = I'll exercise. If I exercise, then I won't eat too much.
--------
If I understand it correctly, the symbolic statement would look like this:
(not p)implies q
=================================
If I exercise, then I won't eat too much.
My answer:
q implies (not p)
----------------------
Truth Table:
p..q:::::q......implies.....not p
t..t:::::t.........F...........f
t..f:::::f.........T...........f
f..t:::::t.........T...........t
f..f:::::f.........T...........t
==========================================
Cheers,
Stan H.
==========================================

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x/30 = 3/15

x*15 = 3*30

15x = 90

x = 90/15

x = 6


So the answer is x = 6

If you need more help, email me at jim_thompson5910@hotmail.com

Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you

Jim

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"The temperature is not 90° and the air conditioner is working, but the apartment is hot." translates to (~p ^ q) ^ r

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Rule for ~    "T after it becomes F; F after it becomes T" 
Rule for /\   "T on both sides of it becomes T, otherwise F"
Rule for \/   "F on both sides of it becomes F, otherwise T"
Rule for ->   "T on left of it and F on right of it becomes F, otherwise T"
Rule for <->  "T on both sides of it or F on both sides of it becomes T, otherwise F"


~(~pΛq)Vr 

p  q  r ~p  (~p/\q)  ~(~p/\q)  ~(~p/\q)\/r  
T  T  T  F     F         T          T
T  T  F  F     F         T          T
T  F  T  F     F         T          T
T  F  F  F     F         T          T
F  T  T  T     T         F          T 
F  T  F  T     T         F          F
F  F  T  T     F         T          T
F  F  F  T     F         T          T


and 

(p→q)↔[(~pΛq)Vp]

p  q  ~p  (p->q)  (~p/\q)  [(~p/\q)\/p]  (p->q)<->[(~p/\q)\/p] 
T  T   F    T        F           T               F
T  F   F    F        F           T               F
F  T   T    T        T           T               T
F  F   T    T        F           F               T


Edwin

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~(p V ~q)

~(p) ^ ~(~q)

~p ^ q

So the two expressions are equivalent

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If 4+6+5=242038 and 3+8+9=242743 then what does 7+9+4= ?

This is an unusual use of an equal sign =, because those
are certainly not equal under the normal use of equal signs.
I will assume you mean that

4+6+5 corresponds to 242038 and
3+8+9 corresponds to 242743

And the question is,

What does 7+9+4 correspond to?

Now 4+6+5 = 15, which corresponds to 242038,
and 3+8+9 = 20, which corresponds to 242743.

So since 7+9+4 = 20 also, it must also correspond to the same number 
which 3+8+9 corresponds to.  So the answer is also 242743.

Edwin

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(q V r)<-->~p


(F V F)<-->~T


(F V F)<-->F


F <--> F


T

So (q V r)<-->~p is true when p is true, q is false, and r is false

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The drop box! I thought I had seen the last of it when my son graduated from Connections Academy.
It is obvious to me that 1 and 2 are equivalent. Statement 3 is not equivalent to them. So the right answer is b).
How to prove it to your teacher's content is another story.
Your lessons or textbook must give you the right format for your proof.
A=it's sunny
B=I go swimming.
Statement 1 says If A, then B.
Statement 2 says If not B, then not A. That is the contrapositive of statement 1, and they are both equivalent.
Statement 3 says Either A or B is true, which I take to imply that both cannot be true at the same time.
They may have taught you ways to express the statements and symbols to abbreviate
the "If ...,then ..." relationship and the "not" in the statements.
They would have told you that a statement implies its contrapositive. If one is true, so is the other.
Alternatively you could have been told to use diagrams to express relationships.
If A, then B can be represented as
The inner circle would represents "A is true" (It's sunny).
Every point in that circle is certainly in circle B.
The large circle,B , represent "B is true" (I go swimming).
I may go swimming when it's not sunny, so those cases would be in the part of circle B that is not inside circle A.
You can graphically see that if not B (if you are outside circle B), not A (you are outside of circle A too.
Either A or B (but not both) , meaning "Either it is sunny or I will go swimming." would be represented by
You cannot be in both circles at the same time.

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 q--->(p v ~q). 
 
p | q | ~q | (p v ~q) |  q-->(p v ~q)
T | T |  F |    T     |       T
T | F |  T |    T     |       T
F | T |  F |    F     |       F
F | F |  T |    T     |       T
 
Rule for ~A     : Put the opposite of what A is. 
Rule for A v B  : Put T except when A and B are both F, then put F
Rule for A--->B : Put T except when A is T and B is F, then put F
 
Edwin

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The parts that make up that expression q V (p -> ~r)  are 
p, q, r, ~r, (p -> ~r),  and q V (p -> ~r).  Make a heading for each of these.
When there are three variables, the truth table will have 8 rows.

Start with this:

p | q | r | ~r | (p -> ~r) | q V (p -> ~r)
------------------------------------------
  |   |   |    |           | 
  |   |   |    |           | 
  |   |   |    |           | 
  |   |   |    |           | 
  |   |   |    |           | 
  |   |   |    |           | 
  |   |   |    |           | 
  |   |   |    |           | 

Fill in the p column with the first half T's and last half F's

p | q | r | ~r | (p -> ~r) | q V (p -> ~r)
------------------------------------------
T |   |   |    |           | 
T |   |   |    |           | 
T |   |   |    |           | 
T |   |   |    |           | 
F |   |   |    |           | 
F |   |   |    |           | 
F |   |   |    |           | 
F |   |   |    |           |

Fill in the q column with 2 T's, 2 F's, 2 T's, 2 F's:

p | q | r | ~r | (p -> ~r) | q V (p -> ~r)
------------------------------------------
T | T |   |    |           | 
T | T |   |    |           | 
T | F |   |    |           |
T | F |   |    |           | 
F | T |   |    |           |
F | T |   |    |           |  
F | F |   |    |           |
F | F |   |    |           |

Fill in the r-column alternating T,F,T,F,T,F,T,F 

p | q | r | ~r | (p -> ~r) | q V (p -> ~r)
------------------------------------------
T | T | T |    |           | 
T | T | F |    |           | 
T | F | T |    |           |
T | F | F |    |           | 
F | T | T |    |           |
F | T | F |    |           |  
F | F | T |    |           |
F | F | F |    |           |

Fill in the ~r column as the opposite of the r column

p | q | r | ~r | (p -> ~r) | q V (p -> ~r)
------------------------------------------
T | T | T |  F |           | 
T | T | F |  T |           | 
T | F | T |  F |           |
T | F | F |  T |           | 
F | T | T |  F |           |
F | T | F |  T |           |  
F | F | T |  F |           |
F | F | F |  F |           |

Fill in the (p -> ~r) column by this rule:  If the
p column has a T and the ~r column has a F, then put
an F in the (p -> ~r) column; otherwise put a T.

p | q | r | ~r | (p -> ~r) | q V (p -> ~r)
------------------------------------------
T | T | T |  F |     F     | 
T | T | F |  T |     T     | 
T | F | T |  F |     F     |
T | F | F |  T |     T     | 
F | T | T |  F |     T     |
F | T | F |  T |     T     |  
F | F | T |  F |     T     |
F | F | F |  F |     T     |

Fill in the q V (p -> ~r) by this rule:
If both the q column and the (p -> ~r) column
have F's, then put an F in the q V (p -> ~r)
column; otherwise put T.

p | q | r | ~r | (p -> ~r) | q V (p -> ~r)
------------------------------------------
T | T | T |  F |     F     |      T 
T | T | F |  T |     T     |      T
T | F | T |  F |     F     |      F
T | F | F |  T |     T     |      T
F | T | T |  F |     T     |      T
F | T | F |  T |     T     |      T
F | F | T |  F |     T     |      T
F | F | F |  F |     T     |      T

Edwin

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T	F	T
F	T	T

p or not p is always true.

John

My calculator said it, I believe it, that settles it

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you might want to check your interpretation...

"A tree IS 4.0m high" ___ "how high will it be in 5 years?"

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p	q	~p	~q	~p V ~q	~(~p V ~q)
T	T	F	F	F	T
T	F	F	T	T	F
F	T	T	F	T	F
F	F	T	T	T	F

Notes:

1) ~p is the opposite of p. So if p is true, then ~p is false (and vice versa). Also, if p is false, then ~p is true. This applies to q and ~q as well.

2) p v q is only false when both p and q are false.

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Since



is a concave up parabola such that the vertex is a minimum point and where the -coordinate of the vertex is equal to and the -coordinate of the vertex is , the minimum value of the function is, indeed, 1 and the given proposition is true.

John

My calculator said it, I believe it, that settles it

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Let s = "My car does not start."
Let w = "I will make it to work on time."

I) If my car does not start, then I will not make it to work on time.

That's ~s->~w

s  w  ~s  ~w  ~s -> ~w
T  T   F   F      T
T  F   F   T      T
F  T   T   F      F
F  F   T   T      T

II) If my car does not start, then I will make it to work on time.

That's ~s -> w
 
s  w  ~s  ~s -> w
T  T   F     T
T  F   F     T
F  T   T     T
F  F   T     F


III) Either my car starts or I will make it to work on time.

That's s V w

s  w  s V w
T  T    T
T  F    T    
F  T    T    
F  F    F    

II and III are equivalent because they have the same truth table,
TTTF.  

Or you can prove it because 

(p->q) <=> (~pVq) 

Let p = ~s and q = w, then

(~s->w) <=> (~(~s) V w)

(~s->w) <=> (s V w)

Edwin

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(pvq)^[p^q]

(pvq)^[q^p]         commutative 

    [(pvq)^q]^p     associative    

[(pvq)v(q^q)]^p     distributive  

    [(pvq)vq]^p     idempotent

[(p^q)^p]v(q^p)     distributive

[(q^p)^p]v(q^p)     commutative

[q^(p^p)]v(q^p)     associative

    [q^p]v(q^p)     idempotent

       q^p          idempotent

Edwin

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Only 1 Liar:

If A is the Liar, then both P and B must be telling the truth. But P and B made contradictory statements. A cannot be the Liar.

If P is the Liar, then A points the finger at P and this is not contradicted by assuming B and T were honest.

If T is the Liar, then again, P and B are contradictory. T cannot be the Liar.

If B is the Liar, then P and A are contradictory. B cannot be the Liar.

P did it.

One Truth Teller:

If A told the truth, P and B are again contradictory. A not the Truth Teller.

If P told the truth, then T must be lying, but B and T cannot both be the perpetrator, hence P is not the Truth Teller.

If T told the truth, then P and B are contradictory again. T is not the Truth Teller

If B told the truth, then A lied so P didn't do it and P lied so B didn't do it (which is what B said) and T lied and therefore T did it.

John

My calculator said it, I believe it, that settles it

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No conclusion.
There is only one implication and the 2nd statement doesn't meet its condition.

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here's your truth tables.


the first truth table is for (pvq) -> (p^q)
you have 2 variables so you need 4 rows (2 * 2).
all possible conditions are shown in the truth table for the 2 variables.
(pvq) is true except when both p and q are false (FF).
all other conditions make it true (TT, TF, FT).
(p^q) is false except when both p and q are true (TT).
all other conditions make it false (TF, FT, FF).
(pvq) -> (p^q) is true except when (pvq) is true and (p^q) is false (TF).
all other conditions make it true (TT, FT, FF).

the second truth table is for (p->q) <-> ~r
you have 3 variables so you need 8 rows (2 * 2 * 2)
all possible combinations are shown in the truth table for the 3 variables.
(pvq) -> (p^q) is true except when (pvq) is true and (p^q) is false (TF).
all other conditions make it true (TT, FT, FF).
~r is true when r is false and ~r is false when r is true.
(p->q) <-> ~r is true when the truth tables for (p->q) and ~r are the same TT, FF).
(p->q) <-> ~r is falser when the truth tables for (p->q) and ~r are not the same (TF, FT).

in general, this is how the truth tables work.

A or B is true in all cases except when A and B are false.
A and B is false in all cases except when A and B are true.
A implies B is true in all cases except when A is true and B is false.
A if and only if B is true when A and B agree (either both true or both false) and is false when A and B disagree (one is true and the other is false).
not A is true if A is false and is false if A is true.

here's a truth table that shows all the possible combinations.

           A   B   ~A    A or B    A and B    A implies B    A if and only if B
           T   T    F       T         T            T                   T
           T   F    F       T         F            F                   F
           F   T    T       T         F            T                   F
           F   F    T       F         F            T                   T

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The base cases (n=1, 2) hold. For some n>2, if

, then



from our original inequality, we establish





Hence, and the induction is complete.

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1+5+9+...+(4n-3) = n(2n-1)

Note: The last term is (4n-3)

-------------------
It is only necessary to show the formula works
for n=1 before showing that if it works for n = k
it works for n = k+1  but I will show it works for
1, 2 and 3.

It works for n=1 because

Last term = 4n-3 = 4*1-3 = 4-3 = 1

1 = 1(2*1-1) 
1 = 1(2-1)
1 = 1(1)
1 = 1

It works for n=2 because

Last term = 4n-3 = 4*2-3 = 8-3 = 5

1+5 = 2(2*2-1) 
1+5 = 2(4-1)
1+5 = 2(3)
1+5 = 6

It works for n=3 because

Last term = 4n-3 = 4*3-3 = 12-3 = 9

1+5+9 = 3(2*3-1) 
1+5+9 = 3(6-1)
1+5+9 = 3(5)
1+5+9 = 15

Now we assume that k is some integer for which
the formula works (and we have 3 such values 
already for which we know the formula works.)

So we assume

1+5+9+....(4k-3) = k(2k-1).

What we want to show is that if we substitute k+1 for n in

1+5+9+...(4n-3) = n(2n-1)

It will also work.

That is we want to show that based on the assumption that it 
works for k, that we'll end up with this without the question
mark over the equal sign:

1+5+9+...+[4(k+1)-3] ≟ (k+1)[2(k+1)-1]
                
The question mark above the equal sign is to show that we
have not shown that yet.  The above is what we must show.

We will simplify what we are to show:

1+5+9+...+[4k+4-3] ≟ (k+1)[2k+2-1]

1+5+9+...+(4k+1) ≟ (k+1)(2k+1)

1+5+9+...+(4k+1) ≟ 2k²+3k+1

Now we will show it.  We start with:

1+5+9+...+(4k-3) = k(2k-1)

We add the k+1st term, which is (4k+1), to both sides,
and hope that the right side comes out to to the
expression above, which is 2k²+3k+1:

1+5+9+...+(4k-3)+(4k+1) = k(2k-1)+(4k+1)
1+5+9+...+(4k-3)+(4k+1) = 2k²-k+4k+1 
1+5+9+...+(4k-3)+(4k+1) = 2k²+3k+1

So we have shown that if the formula works for some
value of n, say n=k, then it will always work for the
next integer n=k+1.

Therefore
Since we have shown it works for n=k=1, it works for n=k+1=2
Since it works for n=k+1=2, it works for n=k+2=3
Since it works for n=k+2=3, it works for n=k+3=4
Since it works for n=k+3=4, it works for n=k+4=5
etc. etc. forever

Edwin

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You could say something like, "For all integers a and b, b^2 is never equal to 4a+2."

For any integer b, b^2 is congruent to either 0 or 1 modulo 4. 4a+2 is always 2 modulo 4. If b^2 and 4a+2 were equivalent mod 4, then they could be equal but they're different mod 4, so they can never be equal.

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Converse:
If we would not....then Washington did not...
Inverse:
If Washington did cross...then we would have...
Contrapositive:
If we won independence... then Washington did cross the Delaware...
John10:)

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your truth table is shown below:

first column is p
second column is q
third column is (p ^ q) which stands for p "and" q.
fourth column is (p v q) which stands for p "or" q.
fifth column is ~(p v q) which stands for "not" (p "or" q).
sixth column is (p ^ q) v (~(p v q)) which stands for:
(p and q) "or" (not (p and q).

The "or" logic is true for all conditions except when both variables involved in the "or" statement are false.
The "and" logic is false for all conditions except when both variables involved in the "and" statement are true.

The first row shows you the headings for each column as expressed in p and q.
The second row shows you the headings for each column as expressed in p, q, A, B, or C.

A is equal to (p ^ q).
B is equal to (p v q).
C is equal to ~(p v q).
The last column shows you (A v C) which translates to (p ^ q) v (~(p v q)).

The second row is not necessary, but i included it to show you that you can set another variable equal to a complex statement to make the statement more readable. If it confuses you, then just skip the second row and go with the headings in the first row as if the second row wasn't there.

The truth table has 4 rows to show all possible conditions for 2 variables.

The are 2 possible conditions for each variable involved. Since there are 2 variables involved, there are 2 * 2 = 4 possible conditions.

~(p v q) is the inverse of (p v q)
if a variable is true, then "not" that variable is false.
if a variable is false, then "not" that variable is true.
In the table, when (p v q) is true, then ~(p v q) is false, and vice versa.

You can put this solution on YOUR website!
your truth table is shown below:




(p v q) is the or statement
(p v q) -> ~p is the implies statement

the or statement is only false if both p and q are false.

the implies statement is only false if (p v q) is true and ~p is false.

v means or
-> means implies
~ means not

You can put this solution on YOUR website!
your truth table is shown below:



v is the or symbol.
~ is the not symbol.
-> is the implies symbol.

the or statement is ~p v q
the implies statement is ~q -> (~p v q)

the or statement is only false if both ~p and q are false.
the implies statement is only false if ~q is true and (~p v q) is false.

there are 4 possible conditions in the truth table because there are 2 variables involved. those variables are p and q. the possible conditions are 2 * 2 (2 possible conditions for each variable used).

You can put this solution on YOUR website!
i)if the dog wags its tail, then the dog is not calm.
ii)either the dog does not wag its tail or the dog is not calm
iii) if the dog is not calm, then the dog wags its tail.

p = the dog wags its tail.
q = the dog is not calm.

~p = the dog does not wag its tail.
~q = the dog is calm.

statement 1 translates to p -> q
statement 2 translates to ~p v q
statement 3 translates to q -> p

statement 1 and 3 are not equivalent
we need to test with statement 2 to see if it's equivalent to either one of the other 2.

when in doubt, you need to do a truth table.
it tells you whether the statement are equivalent or not.

the truth table for these 3 statements is shown in the following picture.

in this table:

statement 1 becomes statement A.
statement 2 becomes statement B.
statement 3 becomes statement C.



The truth table says that (p -> q) is equivalent to (~q v p).

That means that statement 1 is equivalent to statement 2.

It would be difficult to see this without use of the truth table.

Logically, trying to make sense of it takes a little stretch.

I believe the answer lies in the law of disjunction.

The law of disjunction states:

if A or B is true, then:
If A is false, then B must be true.
If B is false, than A must be true.

we have 2 statements that we think are equivalent because the truth table test of equivalency says that they are.

The 2 statements are:

i)if the dog wags its tail, then the dog is not calm.
ii)either the dog does not wag its tail or the dog is not calm

the contra-positive of the first statement and the first statement are equivalent.

The first statement is:
if the dog wags its tail, then the dog is not calm.
The contra-positive to the first statement is:
if the dog is calm, then the dog does not wag its tail.

The second statement is:
either the dog does not wag its tail or the dog is not calm.

If we assume that the dog does not wag its tail is false, then the law of disjunction says that the dog is not calm must be true.

we can write this another way to say:
if the dog wags its tail, then the dog is not calm.
that looks very much like the first statement.

if we assume that the dog is not calm is false, then the dog does not wag its tail must be true by the same law of disjunction.

we can write this another way to say:
if the dog is calm, then the dog does not wag its tail.
that looks very much like the contra-positive to the first statement.

i'm satisfied that the truth table is telling the truth and that statements 1 and 2 are equivalent even though anybody would be hard pressed to determine that they are without using the rules of logic to confirm that.

the truth table test is the way to determine the equivalency of two statements.

that's called demorgan's law i believe.

here's some additional reading for you if i didn't send it to you earlier.
there are references in there as well so you can study to your heart's content if you so desire.

STATEMENT LAWS AND FALLACIES