How to prove A → (B ∨ C) given A → B

Question

How to prove A → (B ∨ C) given A → B

I know this is a valid argument, I'm just terrible at fitch-style proofs and have no idea how to start, let alone finish.

Answer 1

I'm just terrible at fitch-style proofs and have no idea how to start, let alone finish.

Use a conditional proof to introduce that conditional statement. Assume $A$ and then use the premise to derive the conclusion of $B\lor C$.   I do believe you can fill in the rest of the details.

$\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}}$ $$\fitch{1.~~A\to B}{\fitch{2.~~A}{3.~~:\\4.~~B\lor C}\\5.~~A\to(B\lor C)\qquad\text{Conditional Introduction (2-4)}}$$

Answer 2

Use truth table.

A B C A->B A->(BVC)
0 0 0  1     1
0 0 1  1     1
0 1 0  1     1
0 1 1  1     1
1 0 0  0     0
1 0 1  0     1
1 1 0  1     1
1 1 1  1     1

You can see that when A $\rightarrow$ B is true, A$\: \rightarrow$(B$\: \vee \:$C) is true.

Or A$\: \rightarrow$(B$\: \vee \:$C)=(A$\rightarrow$B)$\: \vee\:$C $\Rightarrow$ A$\rightarrow$B.

Answer 3

This result is known as 'Weakening the Consequent'.

$\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}}$ $$\fitch{1.~~A\to B\qquad \text{Hypotheses}}{\fitch{2.~~A \qquad \text{Introduction/Conditional Proof}}{3.~~B \qquad \text{Modus Ponens 1, 2} \\4.~~B\lor C \qquad \text{Disjunction Introduction 3}}\\5.~~A\to(B\lor C)\qquad\text{Conditional Proof(2-4)}}$$

Answer 4

$$A\to B$$

$$\lnot A\lor B$$

$$(\lnot A\lor B)\lor C$$

$$\lnot A\lor(B\lor C)$$

$$A\to(B\lor C)$$

Is this the kind of proof you want?