Is the untested/empty statement true?

Question

I stumbled upon this logic question in a math class recently.

My teacher told us that a statement that is not tested/is empty is true. For example, that if I stated that: "if the team A wins the game, I am gonna buy you a coke", and then team B goes on and wins the game, the statement would be true, independent of me buying a coke. Could anybody elaborate how this can be the case, and why?

It came up as an explanation to why the the empty-set is both an open and a closed set.

Answer 1

The short answer is, this is true because we have adopted that resolution by convention, because, for all its faults, doing so comes closest to resembling the intuitive notion of 'if-then' while having a definite logical value in all mathematical situations. To see what I am getting at, consider the following 'truth table':

A | B | A-->B

T | T | T

T | F | F

F | T | *

F | F | *

This illustrates the truth about our intuitive notion of 'if-then' - When A is true, B has to be true, and when A is false we don't care. But mathematical logic doesn't allow 'we don't care', so we have to replace the '*'-s with definite truth-values. Then the question becomes: how do we do so and still match our intuition? There are clearly four possible resolutions:

A | B | 1 | 2 | 3 | 4

T | T | T | T | T | T

T | F | F | F | F | F

F | T | F | F | T | T

F | F | F | T | F | T

If we choose (1) to mean 'if-then', we are saying 'if A is true then B is true' means 'A and B are both true', and we are also implying 'if B is true then A is true'. But this is clearly not what we want: When we have both 'A-->B' and 'C-->B', this would be saying 'A, B, and C' are all true together'; this doesn't work because sometimes we need A and C to be mutually-exclusive (typical example: a number-theory proof-by-cases, where A means 'n is even' and C means 'n is odd'). A similar problem occurs if we choose (2), which says 'A and B are both true -or- both false'. If we choose (3) to mean 'if-then' we are saying 'if A is true then B is true' means merely 'B is true' - REGARDLESS of A. Clearly that also doesn't work - generally we want B to depend on A somehow. So all that is left is (4).

Answer 2

Do you understand that the conditional $p\rightarrow q$ is true whenever $p$ is false, or whenever $q$ is true?

I think the best way of representing the truth of a condition $p\rightarrow q$ is knowing that $p\rightarrow q$ IS TRUE, UNLESS both $p$ is true, and $q$ is false.

With two variables, $p, q,$ there are four possible assignments of truth values, each represented below.

For example, let's say you promise $p$: = Team A wins the game. Let $q$: = "I'll buy you a Coke."

First of all, you never made any promise about what you'll give me if Team A loses. So if $\lnot p$, I can't accuse you of making a false statement, whether you give me a Coke or not. And in the case that team A wins and you buy me a Coke, well, you've made good on your promise.

The only way you'd be lying is if team A wins ($p$), and you don't buy me a Coke $(\lnot q).$

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Another classic example.

The empty set is a subset of every set.

Let's simply look at some arbitrary set $A$: If any set $B$ is a subset of $A$, then we know by definition that $b\in B \rightarrow b \in A$. Suppose now that $B =\varnothing$. Well, $\varnothing \subseteq A$ because if $b \in \varnothing,$ then $b\in A$. Well, there's no $b\in \varnothing$, but the definition still holds because "$b \in \varnothing$ is false, which makes $b \in \varnothing \rightarrow b \in A$ true

Answer 3

Without going into a formal treatment, what your teacher means is:

The statement $$\forall x\in X: A (x) \Rightarrow B (x) $$ is true if $A (x) $ is false for all $ x\in X$ , no matter what $ B $ is.

That your teacher is right follows from the DEFINITION of the right arrow $\Rightarrow $, that is, it follows from the fact that the truth table of this symbol always assigns "True" when the first argument is false.

There is no epistemological claims behind what your teacher said.

Answer 4

The general idea at which you are hinting is that of conditional probability, which states that for two events, $P$ and $Q$, the statement $P$ implies $Q$ is given by

$$(P \implies Q) \iff (Q \lor \lnot P),$$

where "$\implies$" denotes the implication operator.

To better understand this, consider the following timely scenario. A politicians states: $$\textrm{"If } \underbrace{\textrm{I win the elections, }}_{\textrm{Statement } P} \textrm{then } \underbrace{\textrm{taxes will go down"}}_{\textrm{Statement } Q}.$$ This statement would have a truth value of $T$ is he/she does not lie, and a truth value of $F$ if he/she does lie.

• If $P$ is $T$, and $Q$ is $T$, then the politician has not lied; he/she was elected, and taxes went down. Hence $P \implies Q$ is $T$.
• If $P$ is $T$, and $Q$ is $F$, then the politician has lied; he/she was elected, and taxes did not go down. Hence $P \implies Q$ is $F$
• If $P$ is $F$, regardless of what $Q$ is, the politician cannot lie; he/she was never elected, and so never got the opportunity to lie. Hence $P \implies Q$ is $T$.