SOLUTION: Use an ordinary truth table to answer the following problems. Construct the truth table as per the instructions in the textbook. Given the argument: S ⊃ (K ∨

Algebra ->  Conjunction -> SOLUTION: Use an ordinary truth table to answer the following problems. Construct the truth table as per the instructions in the textbook. Given the argument: S ⊃ (K ∨       Log On


   

Question 915332: Use an ordinary truth table to answer the following problems. Construct the truth table as per the instructions in the textbook.
Given the argument:
S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
This argument is:

a. Invalid; fails in 2nd line.
b. Invalid; fails in 3rd line.
c. Invalid; fails in 1st line.
d. Valid.
e. Invalid; fails in 4th line.
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Question 915332

There was a slight mistake before.  The argument is valid. Here is the corrected version.


I do truth tables different, but it's the same way. I begin
by rewriting the argument with T's and F's under the S's and K's:

ORIGINAL TRUTH TABLE

S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
T ⊃ (T ∨ ∼ T) / T ⊃ T // T ≡ T
T ⊃ (F ∨ ∼ T) / F ⊃ T // T ≡ F
F ⊃ (T ∨ ∼ F) / T ⊃ F // F ≡ T
F ⊃ (F ∨ ∼ F) / F ⊃ F // F ≡ F

Then I place the truth values under the symbols ∼, ⊃ and ≡

S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
T ⊃ (T ∨ F T) / T T T // T T T
T ⊃ (F ∨ F T) / F T T // T F F
F ⊃ (T ∨ T F) / T F F // F F T
F ⊃ (F ∨ T F) / F T F // F T F

Then I erase all the used truth values:

S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
T ⊃ (T ∨ F  ) /   T   //   T 
T ⊃ (F ∨ F  ) /   T   //   F 
F ⊃ (T ∨ T  ) /   F   //   F 
F ⊃ (F ∨ T  ) /   T   //   T 

Place truth values under symbol ∨:

S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
T ⊃ (T T F  ) /   T   //   T 
T ⊃ (F F F  ) /   T   //   F 
F ⊃ (T T T  ) /   F   //   F 
F ⊃ (F T T  ) /   T   //   T

Erase used truth values:

S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
T ⊃    T      /   T   //   T 
T ⊃    F      /   T   //   F 
F ⊃    T      /   F   //   F 
F ⊃    T      /   T   //   T 

Place truth values under ⊃ 

S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
T T    T      /   T   //   T 
T F    F      /   T   //   F 
F T    T      /   F   //   F 
F T    T      /   T   //   T 



Erase used truth values:

S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
  T           /   T   //   T 
  F           /   T   //   F 
  T           /   F   //   F 
  T           /   T   //   T

Place truth values under /, which is the same same 
as ∧ (and, conjunction)

S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
  T           T   T   //   T 
  F           F   T   //   F 
  T           F   F   //   F 
  T           T   T   //   T


Erase the used truth values


S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
              T            T 
              F            F 
              F            F 
              T            T

Place truth values under the // which is the same as ⊃


 S ⊃ (K ∨ ∼ S) / K ⊃ S // S ≡ K
-------------------------------
               T        T   T
               F        T   F
               F        T   F
               T        T   T

Valid.  There are only T's under the //

Edwin

Answer by AnlytcPhil(1806) About Me  (Show Source):