Tutors Answer Your Questions about Proofs (FREE)
Question 981103: Can anybody help me complete this proof?
1. (C v S) v L
2. ~S & (~L v Z)
3. (L v C) > ~Z
/ ~(S v C)
4. ~S Simplification line 2
5. (~L v Z) & ~S Comm line 2
6. ~L v Z Simplification line 5
I think I'm on the right track, could anybody assist me? Thank you
Click here to see answer by jim_thompson5910(35256)  |
Question 981871: 1. H v (~T > R)
2. H v (E > F)
3. ~T v E
4. ~H & D / R v F
5. ~H Line 4 Conjunction
6. (~T v R) Lines 1 and 5 by Disjunctive Syllogism)
Thanks!
"Basic Outline: Free up ~H. Then free up (~T > R) and (E > F). Conjunct them together, and use constructive dilemma."
Hi Jim, would you be able to show me the steps? I'm having some trouble with it, thanks!! :)
Click here to see answer by jim_thompson5910(35256) |
Question 981975: 1. Q > ~Q
2. ~(G & Q) > U
/ ~(G & ~U)
Can somebody solve this proof please? Thank you :)
"You can do a proof by contradiction.
Step 1) assume the complete opposite of the conclusion. Assume (G & ~U)
Step 2) Use simplification to get ~U
Step 3) Use modus tollens with line 2 and ~U to get ~~(G & Q) which turns into (G & Q)
Step 4) Simplification frees up Q
Step 5) Modus ponens on line 1, and using the freed up Q from step 4, consequently frees up ~Q
Step 6) We have Q and ~Q they conjunct to (Q & ~Q) which is always false. This is a contradiction.
Since we have a contradiction, the initial assumption (G & ~U) is false which makes the opposite true. That proves ~(G & ~U) is a proper conclusion."
Jim, is there anyway to do this proof without using a contradiction and not assuming? Thank you :)
Click here to see answer by jim_thompson5910(35256) |
Question 984556: Hello,
I'm having an issue with a problem. (the stars signify the dots)
Regular Proof:
1.~E
2.~(E*D)>F
3.(~FvB)*(~FvC) / Av(B*C)
As well as this one too:
Regular Proof:
1. ~ Bv(B>~A)
2. (B>~A)>(~C*D)
3.~B>~C / C>~C
This ~ is a tilde and these > represent "horseshoe"
Click here to see answer by solver91311(24713)  |
Question 986849: Complete the truth table to show whether the following argument is valid or invalid. If the argument is invalid, you must specify a counter-example.
Premise 1: J → (K→ L)
Premise 2: K → (J → L)
Conclusion: (J v K) → L
J K L J → (K → L) K → (J → L) (J v K) → L
T T T
T T F
T F T
T F F
F T T
F F T
F T F
F F F
Click here to see answer by solver91311(24713) |
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