Implications and open sentences

Consider the statement:

For all \(x\in\mathbb{Z}\), if \(x\) is divisible by \(6\) then \(x\) is even.

“\(x\) is divisible by \(6\)” is an open sentence \(P(x)\)
“\(x\) is even” is an open sentence \(Q(x)\)
“if \(x\) is divisible by \(6\) then \(x\) is even” is an open sentence \(P(x)\implies Q(x)\).

Analysis of quantified implication

“For all \(x\in\mathbb{Z}\), if \(x\) is divisible by \(6\) then \(x\) is even”

is a statement that “ands” together \(P(x)\implies Q(x)\) as \(x\) runs over the integers:

\[\cdots(P(-5)\Rightarrow Q(-5))\wedge(P(-4)\Rightarrow Q(-4))\wedge\cdots\wedge(P(3)\Rightarrow Q(3))\cdots\]

This will be true if every one is true, meaning one of the following is true:

\(x\) is not divisible by \(6\), so \(P(x)\) is false for that \(x\)
\(x\) is divisible by \(6\) and \(x\) is even meaning \(P(x)\) and \(Q(x)\) are true for that \(x\).

It will be false if there is at least one \(x\) that is divisible by \(6\) but not even.

It is a common convention to read a statement like

“If \(x\) is an integer divisible by \(6\), then \(x\) is even”

as including an implicit quantifier “for all \(x\in\mathbb{Z}\).”

If \(f:\mathbb{R}\to\mathbb{R}\) is a differentiable function, then it is continuous
If \(x\in\mathbb{R}\), then \(x^2= x\) implies \(x=0\) or \(x=1\).