Let $a,b,c\in \mathbb Z_{>0}$. Show that $c\cdot\gcd(a,b)=\gcd(ac,bc).$

Consider $d\in\mathbb Z$ such that $d\mid a$ and $d\mid b$. We have $m,n\in\mathbb Z$ such that $a=md$ and $b=nd$. This gives $ac=mdc$ and $bc=ndc$. Therefore $cd\mid ac$ and $cd\mid bc$. We now have to show that the common divisors of $ac$ and $bc$ are the same numbers as $c\cdot\gcd(a,b)$. Consider $d'\in\mathbb Z$ such that $d'\mid ac$ and $d'\mid bc$. We have $q,r\in\mathbb Z$ such that $ac=qd'$ and $bc=rd'$. But from here on I can't continue?

@Semiclassical This means that there exists a statement after A which is true. We will take B to be true for this example.
Now since B is true C must = false. Which means we once again have a negated statement. Hence the next statement is true. However; earlier we said that D would be false. In this case D=1=0. Which results not in a negated statement but a statement that is always 0. (Regardless of what it contains).
So to sumarise so far:
A is negated ($\lnot A)$, B is true (B) and C is negated ($\lnot C)$ while D is always false (0).

@Semiclassical Adding a new statement (e) we will see a similar pattern emerge. What occurs is e begins it life as just e, then due to the statement C it becomes $\bar e$. (D has no effect beucause D is always false and the statement is in effect 'gone'). B ensures e must be negated as well.

the next statement (f) to be introduced must be true becuase (e) ensures that it is. Howver the earlier statements make it false. Hence e=1=0. Which means e is a statement that is always 0. (like D). THe next element will be negated, and the one after it 0, and so on.

@Semiclassical This may be a bit easier to understand:
It should be along the lines of this:
The first is false. By false I mean S=0. Since S=0 is equivalent to S(0)+$\bar S$(1). Setting S to 0 in that manner (S=0) makes the statement that was S become NOT S ($\bar S$)

Next because the first is false the next statement is true. Which means that everything after it is false.

The third statement must be false as per the second. When we make a statement false; we perform the S=0 operation. This results in the statements negation.