How to prove $A \to (B \lor C)$ therefore $(A \to B) \lor (A \to C)$?

Question

In doing this proof I found a solution, but I believe it to be incorrect because within the proof it uses
assume A
.
.
. . . assume C
. . . A -> C
Is it valid to conclude A -> C in the same subproof where you assume C? Also, is this a named equivalence?

Answer 1

Suppose otherwise. Then $A\to(B\lor C)$ while $\lnot((A\to B)\lor(A\to C))$, so that $\lnot(A\to B)$ and $\lnot(A\to C)$. Then $\color{red}{A}$ and $\color{green}{\lnot B}$ from the former and $\color{red}{A}$ and $\color{blue}{\lnot C}$ from the latter. But then $A\to(B\lor C)$ gives either $\color{red}{\lnot A}$, a contradiction, or $B\lor C$, but that gives either $\color{green}{B}$, a contradiction, or $\color{blue}{C}$, another contradiction.

Answer 2

Is it valid to conclude A -> C in the same subproof where you assume C?

Yes. After all, if something is considered true inside a context, it will still be considered true under further assumptions - by rule of "Reiteration".

This also includes any context raised by assumption that the thing is true.

Also, is this a named equivalence?

No. Rather, it is a trivial sub proof involving the Rules of Inference: "Assumption", "Reiteration", and "Conditional Introduction" rules.

Although some proof systems will allow you to immediately derive $A\to C$ under the assumption of $C$ through a version of Conditional Introduction.

Please refer to your system's rule set.

In short: A conditional will be valued as true whenever its consequent is valued as true; so $C$ will entail $A\to C$.

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These parts of your proof will be quite valid. (PS: Yes, you may repeat an assumption.)

$\def\fitch#1#2{~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{A\to(B\vee C)}{\fitch{\text{maybe some other assumptions}}{\fitch{A\hspace{16ex}\text{Assumption}}{~~\vdots\\\fitch{C\hspace{14ex}\text{Assumption}}{\fitch{A\hspace{12ex}\text{Assumption}}{C\hspace{12ex}\text{Reiteration}}\\A\to C\hspace{9ex}\text{Conditional Introduction}\\~~\vdots}\\~~\vdots}\\~~\vdots}\\~~~~\vdots\\(A\to B)\vee (A\to C)}$

So, if you fill in the remainder with valid inferences and sub-proofs, you will have a valid proof.

Answer 3

Short version: If you want a worked solution, then you will find it on pp. 8 to 9 of https://www.logicmatters.net/resources/ifl2/Exercises_solutions_22.pdf

Long version: this is tricky to prove in a standard Fitch-style natural deduction system. But the result isn't constructively valid so you know that you are going to have to use the double negation rule somewhere (the rules for the conditional and the disjunction alone won't suffice).

And indeed, it is a good rule of thumb that if you want to prove a disjunction and haven't a disjunctive premiss to hand, then assume the negation of that target disjunction, prove a contradiction, and so show that the negation of the negated disjunction holds -- and finally appeal to DN to get rid of the double negation.

Shameless advertising: My intro logic text is freely available from https://www.logicmatters.net. Both the exercises and very extensive worked answers to exercises are separately available online too. Worth checking out in particular for natural deduction exercises and the worked answers give many tips for proof-finding.