If $a, b$ and $c$ are positive integers with $a, b ≥ 2$, then equation (1.1) has at most one solution in positive integers $x$ and $y$ with $b^y ≥ 6000 c^{1/δ∗(a,b)}.$
I'm unclear about the above usage of the word 'with'; is this theorem saying
IF $b^y ≥ 6000 c^{1/δ∗(a,b)}$ THEN equation (1.1) has at most one solution
or
IF equation (1.1) has at most one solution THEN $b^y ≥ 6000 c^{1/δ∗(a,b)},$
or
is it an IFF statement?
If $a, b$ and $c$ are positive integers with $a, b ≥ 2$, then equation (1.1) has at most one solution in positive integers $x$ and $y$ with $b^y ≥ 6000 c^{1/δ∗(a,b)}.$
In the above theorem, the two copies of with both have a logical 'and' sense and can be replaced with the phrase 'such that':
If $a,b$ and $c$ are positive integers such that neither $a$ nor $b$ is $1,$ then if $x$ and $y$ are positive integers such that $b^y ≥ 6000 c^{1/δ∗(a,b)},$ then equation (1.1) has at most one solution
$a,b$ and $c$ are positive integers and neither $a$ nor $b$ is $1\quad\implies$
($x$ and $y$ are positive integers and $b^y ≥ 6000 c^{1/δ∗(a,b)}$ $\implies$ equation (1.1) has at most one solution)
$a,b,c,x$ and $y$ are positive integers and neither $a$ nor $b$ is $1$ and $b^y ≥ 6000 c^{1/δ∗(a,b)}\quad\quad\implies$ equation (1.1) has at most one solution.
is this theorem saying
IF $b^y ≥ 6000 c^{1/δ∗(a,b)}$ THEN
No. That implicit conditional (the second '⟹' in the second bullet) above is actually not due to the word with per se, because its condition is actually the entire chunk “in positive integers $x$ and $y$ with $b^y ≥ 6000 c^{1/δ∗(a,b)}\,$”.
In this answer, I explain why the words with and where are frequently ambiguous in logic translations.
