| m | a | m → a |
¬(m → a) |
|---|---|---|---|
| T | T | T | F |
| T | F | F | T |
| F | T | T | F |
| F | F | T | F |
$m$ := the monkey dances
$a$ := I get an A on the test
$Q$ := the monkey dancing IS SUFFICIENT for me to get an A on the test
$\lnot Q$ := the monkey dancing IS NOT SUFFICIENT for me to get an A on the test
It makes sense that I can translate the statement m → a to the statement $Q,$ because if $m$ is false I can understand why this statement is vacuously true, while if $m$ is true then $a$ is true. And it makes sense that its negation ¬(m → a) is equivalent to the statement $\lnot Q.$
But it doesn't make sense that $m$ and $a$ both being true should contradict $\lnot Q.$ The monkey dancing together with me getting an A doesn't actually disprove the monkey dancing is NOT SUFFICIENT for me to get an A on the test, since I could have gotten an A regardless of whether the monkey danced. In this case, shouldn't $\lnot Q$ be vacuously true (contrary to what the truth table says) in the same way that a conditional with a false premise is vacuously true? Or does $m$ and $a$ both being true make $\lnot Q$ just "vacuously false"?