Natural Deduction $(\phi \wedge \psi) \to \chi \vdash (\phi \to \chi) \vee (\psi \to \chi)$

Question

I have to proof the following using natural deduction: $(\phi \wedge \psi) \to \chi \vdash (\phi \to \chi )\vee (\psi \to \chi)$. I want to try to do this without using morgan's laws or anything like that - just the rules for introducing/eliminating the symbols. Can anyone help to get me started?

Isn't what you want to prove false? Take $\phi$ is divisibilty by $2$, $\psi$ divisibility by $3$ and $\chi$ divisibility by $6$, in $\Bbb N$ as a model. Then the left hand holds but neither implication holds. — Henno Brandsma, Sep 14 '20 at 09:38
@HennoBrandsma I see what you mean - but creating a truth table for both sides works, they are both the same. So no I am pretty sure the statement is true — bigbuddies, Sep 14 '20 at 09:40
@HennoBrandsma Ok this is interesting, because I do agree with what you are saying. How come that the truth tables are the same, but the statement is false? Or am I just working on a kind of logic where that doesn't matter — bigbuddies, Sep 14 '20 at 09:46
@HennoBrandsma But for any number $n$ if it is not divisible by 6 one of the following must occur. Either it is not divisible by 2, in which case $\phi \to \chi$ holds, or it is not divisible by 3, in which case $\psi \to \chi$ holds. — Mark Kamsma, Sep 14 '20 at 09:57
It’s not true in intuitionism so we need negation based rules like RAA or TND. — Henno Brandsma, Sep 14 '20 at 10:31

Mauro ALLEGRANZA · Accepted Answer · 2020-09-16T07:26:21.843

3

Improved proof, thanks to comments received.

The proof uses Double Negation rule: thus, it is not intuitionistically valid.

$(\phi \wedge \psi) \to \chi$ --- premise
$\lnot [(\phi \to \chi )\vee (\psi \to \chi)]$ --- assumption [a]
$\phi$ --- asumption [b]
$\psi$ --- assumption [c]
$(\phi \wedge \psi)$ --- from 3) and 4) by $\land$intro
$\chi$ --- using 5) and 1) by $\to$-elim
$(\psi \to \chi)$ --- from 4) and 6), discharging [c]
$\bot$ --- from 7) using $\lor$-intro and 2)
$\chi$ --- from 8) using EFQ
$(\phi \to \chi)$ --- from 3) and 9), discharging [b]

Now we have again a contradiction with 2) and we conclude by DN with:

$(\phi \to \chi )\vee (\psi \to \chi)$ --- discharging [a].

edited Sep 16 '20 at 07:26

answered Sep 14 '20 at 10:08

Mauro ALLEGRANZA

94,169

2

+1. By the way, lines 5, 7, 8 are superfluous. – Taroccoesbrocco Sep 14 '20 at 14:15
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Indeed. You may immediately derive $\chi$ on assuming $\phi$ and $\psi$ (as per line 6), without any need to further assume $\neg\chi$. – Graham Kemp Sep 14 '20 at 23:45
You discharge them as you did, @MauroALEGRANZA . The proof is otherwise okay, it just does not need lines 5, 7, and 8. – Graham Kemp Sep 15 '20 at 21:35

score 2 · Answer 2 · answered Sep 14 '20 at 10:01

As mentioned in the comments, the formula is true (in fact, the converse also holds). This you have already seen in your truth table. However, you ask to prove it without De Morgan's laws, and that is simply impossible. The reason is that it is not a tautology in intuitionistic logic. Put differently: you will need some instance of double negation elimination (or proof by contradiction).

I wrote an answer about this for a more specific case here. They ask about $\neg(P \wedge Q) \to (\neg P \vee \neg Q)$. So setting $\phi = P$, $\psi = Q$ and $\chi = \bot$ we see that that really is a special case of what you are asking about.

Or, for a Kripke frame to show it's not a tautology in intuitionistic logic: let $K := { 0, 1, 1', 2 }$ with $0 < 1 < 2$, $0 < 1' < 2$. And on 0 nothing is true; on 1 just $\phi$ is true; on $1'$ just $\psi$ is true; and on 2, all of $\phi, \psi, \chi$ are true. — Daniel Schepler, Sep 17 '20 at 00:50

Natural Deduction $(\phi \wedge \psi) \to \chi \vdash (\phi \to \chi) \vee (\psi \to \chi)$

2 Answers2