Proof by contradiction (another)

Question

I will suppose that I have to prove a theorem of this type

If a proposition $S$ is verified then; for all $a>0,$ there exists $b>0$ such that the following my properties are satisfying ($p$), ($q$) and ($z$).

to prove this by contradiction I must suppose that there exists $a>0$ , for all $b>0$ we have the negation of $p$ or the negation of $q$ or the negation of $z$

is it equivalent to say

negation of $p$ and property $q$ and property $z$

or

property $p$ with negation of $q$ and $z$

or

property $p$ and property $q$ and negation of $z$

Thank you very much.

Answer 1

If a proposition S is verified then; for all $a>0$, there exists $b>0$ such that the following my properties are satiying (p), (q) and (z).

to prove this by contradiction I must suppose that there exists $a>0$, for all $b>0$ we have the negation of p or the negation of q or the negation of z

Yes.

To prove: $~~~S ~\vdash~ \forall a{>}0~\exists b{>}0~\big(P(a,b)\wedge Q(a,b)\wedge Z(a,b)\big)$

Show that: $~S,~\exists a{>}0~\forall b{>}0~\big(\neg P(a,b)\vee\neg Q(a,b)\vee\neg Z(a,b)\big)~\vdash~\bot$

is it equivalent to say:

• negation of p and property q and property z, or
• property p with negation of q and z, or
• property p and property q and negation of z

No; that would be that exactly one property were false.   You need only show that when at least one property is false, you obtain a contradiction (under the assumption of S).

• negation of p, and
• p but negation of q, and
• p and q but negation of z

Also recall that $(A\vee B)\to C \iff (A\to C)\wedge(B\to C)$ so to show that a disjunction entails a conclusion you need prove each of the disjuncts alone entails the conclusion.