Intuition behind these statements being equivalent

Question

I have encountered a strange equivalence, and I am trying to make intuitive sense of it, which I have been able to do in many similar cases. Here it is:

$$[(p\Rightarrow q)\Rightarrow r] \Longleftrightarrow[((\sim p)\Rightarrow r)\wedge(q\Rightarrow r)]$$

What I'm really looking for is a set of statements in English, for which I will understand that the statements on either side of the biconditional really do say the same thing. What I invented for the statement on the left is this, a description of an electric fence:

$p=$ "I touch the wire"

$q=$ "I get shocked"

$r=$ "There is current in the wire"

Thus: "If my touching the wire is sufficient for my getting shocked, then there is current in the wire." I believe this statement is represented by the expression "$(p\Rightarrow q)\Rightarrow r$".

However, when I try to plug these statements into the expression on the right, I get apparent nonsense: "If I don't touch the wire, then there is current in the wire, and also, if I get shocked, then there is current in the wire." In particular, the first half of this conjunction seems bizarre...

Despite this seeming lack of accord, the truth tables for the two statements show that they are equivalent. In both cases, they are only false in three cases: when the truth values of $(p,q,r)$ are $(T,T,F)$, $(F,T,F)$ or $(F,F,F)$. In all other cases, both statements are true.

Can anyone help me reconcile this strange situation? Is this just the strangeness of material implication rearing its counter-intuitive head, again? Thanks in advance.

Answer 1

I found a nice path from the LHS statement to the RHS statement, and this helps me troubleshoot my example:

$$(p\Rightarrow q)\Rightarrow r \\ (\sim p \vee q)\Rightarrow r\\ \sim(p\,\wedge\sim q)\Rightarrow r\\ (p\,\wedge\sim q)\vee r\\ (p\vee r)\,\wedge(\sim q\vee r)\\ (\sim p\Rightarrow r)\wedge(q\Rightarrow r)$$

This is not the shortest possible path between the statements, but I like that it provides a variety of rephrasings, each close to its neighbors, which make it easy to test example sentences for goodness-of-fit. My electric fence example shows its weakness right away after the first step. The real logical relation there is the other way around:

$$r\Rightarrow(p\Rightarrow q)$$

The current being on is sufficient for the implication: if I touch, I get shocked. The implication "if I touch I get shocked" is necessary for saying the current is on. My mistake was confusing sufficiency and necessity in that situation.

As for a good example for the original statement, based on the answer by @Bertrand Wittgenstein's Ghost, we have:

$p=$ "It rains"

$q=$ "I have an umbrella"

$r=$ "I stay dry"

$p\Rightarrow q=$ "Whenever it rains, I have an umbrella"

Thus:

"If, whenever it rains I have an umbrella, then I stay dry"

and

"If it doesn't rain, then I stay dry; also, if I have an umbrella, then I stay dry."

Answer 2

$(P\rightarrow Q)\rightarrow R)\iff((\lnot P\rightarrow R)\land (Q\rightarrow R ))$

So you set up the propositions like this:

$P:$ I touch the wire

$Q:$ I get shocked

$R:$ There is current in the wire

That would make the left hand side:

If (if i touch the wire, then i get shocked), then there is current in the wire.

Which is the same as: if i touch the wire and then get shocked implies there is current in the wire.

Right hand side would be:

If I don't touch the wire, then there is current in wire and if I get shocked, then there is current in the wire.

Well, think of it like this:

$P:$ Its Raining

$Q:$ I have an umbrella

$R:$ I don't get wet

L.H.S:

If it's raining, and I have an umbrella, then I don't get wet.

R.H.S

If its not raining, then I don't get wet and if I have an umbrella, then I don't get wet.

Both of these statements make intuitive sense. In fact, your initial example makes sense too, however in terms of truth-tables, but if you translate it into colloquial english it is not easy to see because the statement: "If i touch the wire, then I get shocked" assumes there is current in the wire. Keeping this assumption in mind we get: "if I don't touch the wire, then there is current in the wire" obviously, because we assumed there is current in the wire. Similarly, "If i get shocked, then there is current in the wire", obviously because that is our assumption.

See: $(p\rightarrow q)\rightarrow r$ is equivalent to $(\lnot p\lor q)\rightarrow r$

This equivalence shows that $r$ is actually embedded in $(p\rightarrow q)$. That is, if we suppose the $(p\rightarrow q)\rightarrow r$, then we suppose $r$ is is necessarily true.

Aside:

Deduction does not tell us anything new. In fact, it just explicates what other things we implicitly believe given an initial set of beliefs.

hint: Always try to keep your propositions simple.