# How to prove that ∼p → (q ∧ r) is false? [closed]

I am a beginner in logic.

With the premise that (q ∧ r) is false, how can I prove that ∼p → (q ∧ r) is invalid?

This is the last part of a logic problem. I have built this truth table:

p | q | r | ~p | (q ∧ r)
T | T | T | F  |  T
T | T | F | F  |  F
T | F | T | F  |  F
T | F | F | F  |  F
F | T | T | T  |  T
F | T | F | T  |  F
F | F | T | T  |  F
F | F | F | T  |  F

• You can build a truth table and consider all possible values for your variables Aug 15 at 15:00
• Do you know that $p$ is not true? Aug 15 at 15:04
• Yes, there are (p ∨ ∼s) is True, which means p is True and ~s is True. Aug 15 at 15:12
• I do build a truth table, but I'm not sure I do a right thing, now I put my truth table on. Aug 15 at 15:14
• If $p$ is true then the statement is true because "false" implies everything. Aug 15 at 15:15

With the premise that (q ∧ r) is false, how can I prove that ∼p → (q ∧ r) is invalid?

Since $$\color\red{(q ∧ r)}$$ is false, $$(¬p → \color\red{(q ∧ r)})$$ and $$p$$ have the same truth value.

Thus, since $$p$$ can be false, $$(¬p → (q ∧ r))$$ correspondingly can be false, which is what I think you're wanting to prove.

It sometimes true (e.g. when P, Q and R are all true) and sometimes false (e.g.when P, Q and R are all false). It is a contingency. Here is the truth table.

With the premise that (q ∧ r) is false, how can I prove that ∼p → (q ∧ r) is invalid?

Additionally to ryang answer, I will provide a more algebraic approach to this problem.

We use: $$a \rightarrow b \equiv \neg a \lor b$$

The original proposition can be written as

$$\neg p \rightarrow (q \land r) \equiv p \lor (q \land r) \equiv p \lor F \equiv p$$

So, $$\neg p \rightarrow (q \land r)$$ is true if an only if $$p$$ is true.