If $4=5$, then $6=8\,$ (yes or no?) I had an argument with a friend about the statement in the title. I asserted that if $4=5$, then $6=8$, as you can derive any conclusion from a false statement. However, he does not agree, and claims that you cannot know that $6$ would equal $8$ if $4$ were to equal $5$. Is the statement in the title correct?
 A: There is a couple of issues here. First, if a system contains a contradiction then everything is true. So if $4=5$ and $4 \neq 5$ then all statements are true.
Second, implication is defined in such a way that the statement $P \implies Q$ is defined to be true if $P$ is known to be false. This says nothing about $Q.$
 The reason for this is that when $P$ is false, the implication says nothing about $Q.$
A: You have to define the axioms you are working with.  If you are working in PA with $4$ defined as $s(s(s(s(0)))))$ and $5$ defined as $s(s(s(s(s(0)))))$ you are correct as you can derive a contradiction.  In that case, you can derive any statement.  If you are working in basic first order logic with equality, I can define a model of $\{4,6,8\}$ and define $4=5$ but $6 \neq 8$ and your friend wins.  I agree that is perverse as you have presented it.
A: This is an expansion or digression on Ross Millikan's answer. The issue is to determine exactly what you mean by the implication "if $4=5$ then $6=8$".  
One way to read it is as a statement about the real line. Since $4 \not = 5$, the statement "if $4 = 5$ then $6 = 8$" is vacuously true for the real line.
But another way to read an implication like that is as a statement about rings with identity (one could also say it is about semirings). In any ring with identity, a natural number $n$ is naturally represented as a sum $1+1+\cdots +1$ with $n$ ones. I would more naturally read a statement of the form "if $a = b$ then $c = d$", where $a,b,c,d$ are natural numbers, as a statement about all rings with identity, saying that one can derive $c=d$ from $a=b$ using the ring axioms. 
But, in rings with finite characteristic, we may have $n = m$ even though $n$ and $m$ are distinct natural numbers. For example, in the ring $Z_8$ we have $5 = 13$, but in that ring we do not have $6 = 9$. So the implication "if $5 = 13$ then $6 = 9$" is not true if we view it as a statement about all rings. But that statement is true if we view it as a statement about the ring of real numbers. 
In the question, perhaps by luck, you assumed that $4 = 5$, which is true only in the unique trivial ring where $ 1= 0$. It is true that, in that particular ring, that $n = m$ for all natural numbers $n$ and $m$. But that is a special attribute of that ring, not a universal property of all rings with identity. 
A: $$4=5$$
$$4-1=5-1$$
$$2(4-1)=2(5-1)$$
A: Here is an argument that doesn't involve subtleties of logic:


*

*$4=5$

*$4-3=5-3$

*$1=2$

*$2\times 1=2\times 2$

*$2=4$

*$2+4=4+4$

*$6=8$
A: Argument for YES:
If we assume $4=5$, then a contradiction can be derived, and so by the classical logic rule of ex contradictione quodlibet (ECQ), we can indeed infer $6=8$ (by the kind of argument given by vadim123, say). That's because PA (I assume that to be the context of this expression) is a first order (classical logic) theory. That is, $4=5$ entails a contradiction in PA, and consequently everything else, by ECQ; in particular $6=8$.
Argument for NO:
Some argue that not all counterpossibles are trivially true (some can be non-trivially so), and in particular some of them can be false. For example, I suspect that if there existed something like paraconsitent arithmetic, then ECQ wouldn't be valid and so the conditional in question needn't be true. Here is an example of a relevant arithmetic, i.e. based on relevant logic R, rather than FOL. Note: every relevant logic is paraconsistent.
Moreover there are more general theories of (non-trivial) counter-possible reasoning. Here's an answer given to an objection to non-trivial counter-possible reasoning, which I think is relevant to the question you pose. Keep in mind that it stems from a theory of counter-possible reasoning (a theory which posits the existence of non-trivially true counter-possibles, i.e. subjunctive conditionals with impossible antecedents), in this case one developed by Brogaard and Salerno. 
The key thing to note here is that the main assumption is that impossibility comes in degrees. Such degrees are then modeled by ordering impossible situations (worlds) in a manner that the closer ones (to the actual world) are less impossible. Having said that, even if it's not exactly what you were hoping for in terms of an answer, it certainly gives you something to think about.
The excerpt is taken from Francesco Berto's Stanford Encyclopedia of Philosophy article.
"Consider the claim:
(1) If 5 + 7 were 13, then 5 + 6 would be 12.
Prima facie, this is a non-trivially true counterpossible. However, Williamson argues, other non-trivial consequences of the supposition would then be that 5 + 5 = 11, and 5 + 4 = 10, and ... , and 0 = 1. Therefore,
(2) If the number of answers I gave to a given question were 0, then the number of answers I gave would be 1.
but (2) is clearly false.
A reply by Brogaard & Salerno 2007 consists in putting Williamson in front of a dilemma: either we hold the context fixed in this kind of counterpossible reasoning, or we don't. If we don't, then (2) does not follow from (1). In particular, the context at which (2) comes out false is one at which the closest antecedent worlds are possible and, to be sure, at those worlds, 0 is not 1. But if we hold the context fixed, then what does follow is just the following counterpossible:
(3) If 0 were 1 and the number of right answers I gave were 0, then the number of right answers I gave would be 1.
Now this is intuitively true, and non-trivially so."
I suggest reading Berto's article if some of the terms are unclear. To familiarize yourself with the details of Brogaard and Salerno's theory of non-trivial counterpossibles I suggest reading the key articles, which are listed in the Bibliography.
A: $$
4=5
$$
$$
4-4=5-4
$$
$$
0=1
$$
thus 
$$
6=6+0+0=6+1+1=8
$$
A: Vacuously, yes. But for fun you can construct a direct argument that in fact $x = x + 1 \implies \forall y \in \mathbb N. y = x$. We have
$$x = x + 1 \implies 0 = 1 \implies 0 = y - x \implies x = y$$
