1

I'm having a really hard time trying to derive (E v C) from {~A > ~B, A > C, B v D, D > E}. Where '>' is the material conditional, so that 'A > C' is read as "If A then C".

I used negation elimination (A/~E) to assume ~A, and from that got ~B from conditional elimination.

In order to get B, I'm trying to use Disjunction Elimination, but I'm having trouble getting D to yield B.

How do I solve this problem?

2

With Natural Deduction :

1) ¬A → ¬B --- premise

2) A → C --- premise

3) B ∨ D --- premise

4) D → E --- premise

5) ¬A --- assumed [a]

6) B --- assumed [b]

7) ¬B --- from 1) and 5) by →-elimination (modus ponens)

8) --- from 6) and 7)

9) A --- from 5) and 8) by ¬-introduction and Double Negation, discharging [a]

10) C --- from 2) and 9) by →-elimination (modus ponens)

11) E ∨ C from 6) by ∨-introduction

12) D --- assumed [c]

13) E --- from 4) and 12) by →-elimination

14) E ∨ C from 13) by ∨-introduction

We have derived E ∨ C both under assumption B and under assumption D; thus, we can apply ∨-elimination, with premise 3) B ∨ D :

15) E ∨ C from 3), 5)-11) and 12)-14) by ∨-elimination, discharging assumptions [b] and [c].

1

I won't give you a complete answer as this looks like homework, but I'll give you a hint:

You need to look at the possible values of A. What is implied when A is true? And what happens when A is false?

1

{~A > ~B, A > C, B v D, D > E}

We want to show that (E v C) is true.

Now, suppose E is true. Then we're done

So, suppose E is false. Since D > E, D must be false.

Since B v D is true and D is false, B must be true.

Since B is true, ~B is false.

Since ~A > ~B and ~B is false, ~A must be false and A must be true.

Since A > C and A is true, C must be true.

So either E is true, or else C is.

QED. Or as I learned in a MOOC with a lot of French students, CQFD, ce qu'il fallait démontrer.

1

This proof is similar to that provided by user4894 except I presented it in the Fitch-style notation using Klement's proof checker:

enter image description here

The proof uses the following rules: disjunction introduction (vI), modus tollens (MT), disjunctive syllogism (DS), contradiction introduction (⊥I), negation introduction (¬I), double negative elimination (DNE), conditional elimination (→E) and the law of the excluded middle (LEM).

Details about these rules may be found in forall x: Calgary Remix.


References

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/

1

You have B v D as a premise so why not use it? Assuming the case is D, derive C v E from D > E . Assuming the case is B, derive C v E from ~A > ~B and A > C. Eliminate that disjunction.

 1|  ~A > ~B
 2|  A > C
 3|  B v D
 4|_ D > E
 5|  |_ B       Assume
 6|  |  |_ ~A   Assume                :
 7|  |  |  ~B   > elimination 1,6     :
 8|  |  |  #    ~ elimination 5,7     :
 9|  |  ~~A     ~ introduction 6-8    or Modus Tolens 1,5
10|  |  A       ~~ elimination 9
11|  |  C       > elimination 2,10
12|  |  C v E   v introduction 11
  /  /
13|  |_ D       Assume
14|  |  E       > elimination 4,13
15|  |  C v E   v introduction 14
16|  C v E      v elimination 3,5-12,13-15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.