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?

What do you mean by "says x=p only"? Do you mean it only says x=p, or that is says that x=p and that x cannot equal anything else? Additional hypotheses don't ever hurt anything. — MPW, Mar 02 '18 at 18:20
OK, if the question wasn't all that clearly phrased in the beginning, that edit just made it even more confusing! .. what on earth is $x = p \lor q$ as an assumption? I figured that statements were about the possible values of some variable $x$ ... but now $x$ is a logic statement?? — Bram28, Mar 02 '18 at 18:26
Maybe give us the theorem and the assumption and we'll tell you (and hopefully explain) if they contradict each other? — kingW3, Mar 02 '18 at 18:29

score 2 · Answer 1 · answered Mar 02 '18 at 18:20

2

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

answered Mar 02 '18 at 18:20

N. S.

132,525

score 2 · Answer 2 · answered Mar 02 '18 at 18:23

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 $\vee$.

answered Mar 02 '18 at 18:23

Daniel Donnelly

21,342

score 0 · Answer 3 · answered Mar 02 '18 at 18:24

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.

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

3 Answers3