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I am desperately trying to figure out the formal proof for this argument.

$$\begin{array}{r} A\lor B\\ A\lor C\\ \hline A\lor (B \land C) \end{array}$$

I am trying to apply the backwards method here. I am trying to infer A, in order to use vIntro in the last step and introduce the final disjunction. But I got stuck finding sufficient proof for A.

Any hint will be greatly appreciated. Thank you!

johnrookie
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    from $A\vee B$, $A\vee C$, you get $(A\vee B)\land(A\vee C)$ and from this you get $A\lor(B\land C)$. your approach cant work because if $A$ is false and both $B,C$ are true, the prop holds – yoyo Dec 13 '11 at 17:18
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    Well, it isn't really true that you can infer A from what you're given, is it? There is a situation where the assumptions hold, but A is still false, right? What situation is that? What happens then? – Ben FL Dec 13 '11 at 17:07
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    Without knowing what formal proof system you are using, it is impossible to answer. The details of the formal proof will vary immensely from one system to another. – Carl Mummert Dec 13 '11 at 17:20
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    For example, in one system I am used to, every (substitution instance of a) tautology is an axiom. So you can use $(A \lor B) \to (A \lor C) \to (A \lor (B \land C))$ as an axiom, apply modus ponens twice to cut $A \lor B$ and $A \lor C$ and you will be left with the conclusion you want. – Carl Mummert Dec 13 '11 at 17:21
  • If your not familiar with a term like "formal proof system", I'll ask exactly what rules of inference can you use? Exactly, what axioms (if any... you might not have any) can you use? – Doug Spoonwood Dec 13 '11 at 17:51
  • As Doug hints, we don't know what formal proof system you are supposed to use. The course instructor DOES. Why not seek help there? – GEdgar Dec 13 '11 at 20:23
  • Such problems show up here now and then; but it is never defined what is meant by a "formal proof". Usually everyone is happy when the obvious has been regurgitated in the form of ten consecutive sentences in everyday language. – Christian Blatter Dec 13 '11 at 20:35
  • @Christian, "never" is a very strong term. Here is a counterexample. – hmakholm left over Monica Dec 13 '11 at 22:43
  • the OP keeps asking these kinds of (very basic) questions, as has been observed in another comment elsewhere. The problem is that he refuses to provide further information on his deduction method or book reference. He does not even bother to edit his own questions to provide extra information as requested by other members. It certainly looks like a lazy student asking us to do his homework. Prove me wrong (by the "backwards method"). – magma Dec 16 '11 at 13:51

4 Answers4

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Is the following a "formal proof"?

Let $a$, $b$, $c$ be boolean variables representing the truth values of $A$, $B$, $C$. Then by the second distributive law of Boolean algebra we have

$$(a\vee b)\wedge(a\vee c)=a\vee(b\wedge c)\ .$$

This shows that your "argument" not only proves the truth of the third line under the assumption of the first two, but that in fact the stuff above the \hline is logically equivalent to what's underneath.

Christian Blatter
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The following proof uses the law of the excluded middle (LEM) on $A$ to derive the goal:

enter image description here

I consider two cases $A$ and $¬A$. The first case uses disjunction introduction (∨I). The second case uses disjunctive syllogism (DS) and disjunction introduction.

Further information about the inference rules and the proof checker are given in the links below.

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, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf

Frank Hubeny
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I am trying to apply the backwards method here. I am trying to infer A, in order to use vIntro in the last step and introduce the final disjunction. But I got stuck finding sufficient proof for A.

You don't prove it; you assume it -- To be precise: you assume both cases aiming to derive the same conclusion from each. The disjunction elimination rule then discharges the assumptions. This is also known as the Proof By Cases argument. $$\dfrac{\dfrac{[A]^\star\\~~\vdots}{A\lor(B\land C)}\quad\dfrac{[B]^\star\\~~\vdots}{A\lor(B\land C)}\quad\lower{1.5ex}{A\lor B}~}{A\lor(B\land C)}{\small\lor\mathsf E^\star}$$

So, in this proof you need two $\lor$ eliminations and therefore four assumptions must be made and discharged. The trick is how you combine the assumptions of $B$ and $C$.

$$\dfrac{\dfrac{[A]^1}{A\lor(B\land C)}{\small\lor\mathsf I}\quad\dfrac{\dfrac{[A]^2}{A\lor(B\land C)}{\small\lor\mathsf I}\quad\dfrac{\dfrac{[B]^1\quad[C]^2}{B\land C}{\small\land\mathsf I}}{A\lor(B\land C)}{\small\lor\mathsf I}\quad\lower{1.5ex}{A\lor C}}{A\lor (B\land C)}{\small\lor\mathsf E^2}\quad\lower{1.5ex}{A\lor B}}{A\lor (B\land C)}{\small\lor\mathsf E^1}$$

Graham Kemp
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Since you've pointed out the text elsewhere these problems seem to come from, here goes:

1 (A v B) premise
2 (A v C) premise
3 | A hypothesis
4 | (A v (B^C)) 3 V introduction
5 | B hypothesis
6 || A hypothesis
7 || (A v (B^C)) 6 V introduction
8 || C hypothesis
9 || (B^C) 5, 8 ^ introduction
10 || (Av(B^C)) 9 V introduction
11 | (Av(B^C)) 6-7, 8-10, 2 V elimination
12 (Av(B^C)) 3-4, 5-11, 1 V elimination.
Doug Spoonwood
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