## Thinking Mathematically (6th Edition)

RECALL: (1) Modus Ponens states that: $p \longrightarrow q$ $p$ __________________ $\therefore q$ (2) Modus tollens states that: $p \longrightarrow q$ $\neg q$ __________________ $\therefore \neg p$ The given statement does not use modus ponens nor modus tollens. Thus, the argument is invalid. There will be no logical conclusion about $p$ for the conditional $p \longrightarrow q$ when $q$ is known to be true.