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I am trying to prove that the following is a tautology:

$(A \implies (B \implies C)) \implies ((A \implies (C \implies D)) \implies (A \implies (B \implies D)))$

To make progress, I thought I'd eliminate all the arrows. After that, and some de Morgan, I've arrived at:

$(A \land B \land ¬C) \lor (A \land ¬C \land ¬D) \lor (¬A \lor B \lor D) $

At this point, I don't know how to carry on, though. I feel like I'm missing some rule -- I get stuck in trying to expand this and don't get anywhere.

I'd be really grateful for help / hints!

EDIT:

Thank you Henning Makholm and Mauro ALLEGRANZA for spotting mistakes in my reformulations. The rewritten form should read:

$(A \land B \land ¬C) \lor (A \land C \land ¬D) \lor (¬A \lor ¬B \lor D) $

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To use equivalences to find the solution, we will profit from the equivalence scheme: $$\phi\implies(\psi\implies\chi) \quad\equiv\quad \neg\phi\lor\neg\psi\lor\chi$$ for arbitrary formulas $\phi,\psi,\chi$. This can be obtained by applying the rule of material implication $\phi \implies \psi \equiv \neg\phi \lor \psi$ twice.

Applying it to the three innermost formulas, one obtains:

$$(\neg A \lor \neg B \lor C) \implies ((\neg A \lor \neg C \lor D) \implies (\neg A \lor \neg B \lor D))$$

The entire formula, however, also fits our scheme, leading to:

$$\neg(\neg A \lor \neg B \lor C) \lor \neg(\neg A \lor \neg C \lor D) \lor (\neg A \lor \neg B \lor D)$$

Applying De Morgan on the first two entries yields the expression mentioned in your edit: $$(A \land B \land \neg C) \lor (A \land C \land \neg D) \lor \neg A \lor \neg B \lor D$$

How do we go from here? The key is expanding e.g. $\neg B$ to: $$(\neg B \land A \land \neg C) \lor (\neg B \land \neg A) \lor (\neg B \land C)$$

and then regrouping the first term with $(A \land B \land \neg C)$ to yield $A \land \neg C$. We can then continue eliminating the conjunctions with the most terms until we arrive at e.g. a canonical $A \lor \neg A$.


This indicates a way in which you could establish the formula is a tautology with equivalences. It should be clear, however, that this is not a very efficient way to go about it. One would be better off using e.g. the deduction theorem or truth tables.

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There must be something wrong with your rewritings -- what you have arrived at is false when all of the propositional variables are true (as well as when they're all false), whereas the original formula is true in that case.

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  • $\begingroup$ Thank you, I just updated my rewritten version -- hopefully this part is now correct. $\endgroup$ – noctilux Apr 25 '15 at 20:43
  • $\begingroup$ @noctilux: The new rewritten version has the same problem -- as long as each of the conjunctions has both a negated and a non-negated variable, they will all be false in the all-variables-true and the all-variables-false case. So the new rewritten form is not a tautology, and so equivalent to the original one either. $\endgroup$ – Henning Makholm Apr 25 '15 at 22:08
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excuse my unfamiliarity with the formal expression of propositional calculus, but i hope the following reasoning may be of assistance

suppose the proposition false. then we have: $$ (A \to (B \to C)) \land ¬ ((A \to (C \to D) \to (A \to (B \to D))) = \\ (A \to (B \to C)) \land ((A \to (C \to D)) \land ¬(A \to (B \to D)) = \\ (A \to ((B \to C) \land (C \to D)) \land (A \land ¬(B \to D)) = \\ ((B \to C) \land (C \to D)) \land (B \land ¬D) =\\ C \land (C \to D) \land ¬D = \\ D \land ¬D $$

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  • $\begingroup$ $(P\to Q)\wedge P = P\wedge Q$, not just $Q$. $\endgroup$ – user21467 Apr 26 '15 at 12:06
  • $\begingroup$ thx Steven. does that mean my last line should be $C \land D \land ¬D$? $\endgroup$ – David Holden Apr 27 '15 at 15:37
  • $\begingroup$ Yeah. Also the $A$ from a few lines before should survive until the end. $\endgroup$ – user21467 Apr 28 '15 at 0:53
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(If anyone wants a proof by the deduction theorem)

Suppose $(A\implies (B\implies C))$. Then not $A$ or $(B \implies C)$.

Case 1: If not $A$, then $(A⟹(B⟹D))$. So $((A⟹(C⟹D))⟹(A⟹(B⟹D)))$ is true.

Case 2: $A$ is true. So $(B \implies C)$, then not $B$ or $C$.

Case a) If not $B$, then $(A⟹(B⟹D))$ is true.

Case b) $B$. So $C$.

Case bi) If $D$ is false, $C⟹D$ is false, so $(A⟹(C⟹D))$ is false. So $((A⟹(C⟹D))⟹(A⟹(B⟹D)))$ is true.

bii) If $D$ is true, $B⟹D$ is true. So $(A⟹(B⟹D))$ is true. So $((A⟹(C⟹D))⟹(A⟹(B⟹D)))$ is true.

So $(A⟹(B⟹C)) \implies ((A⟹(C⟹D))⟹(A⟹(B⟹D)))$.

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  • $\begingroup$ Hi Ilham, thank you for this. However, I am trying to derive a solution using only equivalences, and furthermore, I am trying to prove that the formula is a tautology. It really is -- see here: wolframalpha.com/input/… $\endgroup$ – noctilux Apr 25 '15 at 21:01
  • $\begingroup$ @noctilux Yeah, I realized it was a tautology that's why I posted this proof. Earlier I had misread the brackets. But you want to use equivalences, so I guess this answer is not applicable to you. Maybe to someone else? $\endgroup$ – Ilham Apr 25 '15 at 21:03

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