Do $p \lor q$ and only $p$ contradict each other?

Question

If a theorem says $x = p$ only, but an assumption gives us $x = p \lor q$. Can I say they contradict each other and thus the assumption is wrong?

Answer 1

($x=p$ only ) and ($x=p$ or $q$) is equivalent to $x=p$ and $x \neq q$.

Answer 2

$(p \vee q) \wedge (p \wedge \neg q) = (p \wedge \neg q ) \vee (p \wedge q \wedge \neg q) = (p \wedge \neg q) \vee F = p \wedge \neg q$

By distributing $\wedge$ over $\vee$.

Answer 3

It depends.

What exactly is meant by '$x = p$ or $q$'?

If it means: 'if $x$ takes on the value of either $p$ or $q$, then $x$ is a solution to the problem', then that does indeed contradict the 'only $x=p$ is a solution' statement ... unless, of course, $p$ and $q$ are actually the same.

But if it means 'the solution is either $x=p$ or $x=q$', then that is perfectly compatible with the statement that only $x=p$ is a solution.