1. A -> B

2. B -> ~A

3. (A v D) v E

4. (D v E) -> F

5. :: A         Conditional Proof (CP) assumption

6. :: B         5,1 Modus Ponens (MP)

7. :: ~A        6,2 MP

8. :: A & ~A    5,7 Conjunction (CONJ), END OF PROOF (contradiction reached)

 

RELATED QUESTIONS

Indirect proof
9. 
1) R
2) (~ C v ~ D) v S
3) ~ (C ⋅ D) ⊃ ~R		/ ∴ S
 (answered by solver91311)

1. A -> ~(B v C)
2. ~D -> (~A -> ~E)
3. ~(~E v F)
4. ~F -> (A -> B)	/D 
I don't... (answered by Edwin McCravy)

1) (A&B) > ~(S v T)
2) ~E
3) (S v T) v ~(~E & ~F)
4) (~E v F) > (A&B)

Conclusion: E  (answered by solver91311)

can i have help solving this proof please?
Construct a regular proof to derive the... (answered by jim_thompson5910)

I need help finishing this proof.
A v (~B v ~C), A -> (D -> E), ~(~B v ~D) :. C -> (D ->  (answered by jim_thompson5910)

INSTRUCTIONS: Construct a regular proof to derive the conclusion of the following... (answered by solver91311)

INSTRUCTIONS: Construct a regular proof to derive the conclusion of the following... (answered by Edwin McCravy)

Using a conditional proof: 

1. (D v E) ⊃ (F · G)

2.( A v B) ⊃ (D · C)    (answered by jim_thompson5910)

Two step proof:
1.	D > E
2.	F > G
3.	D v F	/ E v G
 (answered by Edwin McCravy)