“We have to live today by what truth we can get today and be ready tomorrow to call it falsehood.” - William James
Is the WFF \[ (P\lor \lnot Q) \] true? It depends on what we mean by \(P\) and \(Q\), which depends on the situation!
The technical name for a situation in logic is a model (though sometimes it is called an assignment or an interpretation).
It turns out that in Classical Propositional Logic, if we want to know whether a WFF like \((P\lor \lnot Q)\) is true, we don’t need to know exactly which propositions our propositional variables \(P\) and \(Q\) stand for, e.g.,
All that matters is whether \(P\) and \(Q\) represent true propositions or false propositions.
Definition
A model in Classical Propositional Logic is an assignment of truth values (T or F) to all relevant propositional variables.
Example
If our WFFs contain only the propositional variables \(P\) and \(Q\), then there are four possible models: - \(P\) is true, \(Q\) is false - \(P\) is false, \(Q\) is true - \(P\) is false, \(Q\) is false - \(P\) is true, \(Q\) is true
If the WFFs we care about instead contained variables \(R_1\), \(R_2\), and \(R_3\), then there would be eight (\(2^3\)) possible models.
We cannot say that a WFF like \((P\land Q)\) is true or false until we specify the model.
In Classical Propositional Logic, the truth of a formula is determined by the standard truth table operations.
Propositions of the form \(({\cal A}\land {\cal B})\) (pronounced “\(\cal A\) and \(\cal B\)”) are called conjunctions.
Definition
A WFF of the form \(({\cal A}\land {\cal B})\) is true in a model when both \(\cal A\) and \(\cal B\) are true in the model:
| \(\cal A\) | \(\cal B\) | \(({\cal A}\land{\cal B})\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
Example
In the model
\(P\) is true, \(Q\) is false, \(R\) is true
Propositions of the form \(({\cal A}\lor {\cal B})\) (pronounced “\({\cal A}\) or \({\cal B}\) ”) are called disjunctions.
Definition
A WFF of the form \(({\cal A}\lor {\cal B})\) is true in a model when either \({\cal A}\) and \({\cal B}\) or both are true.
| \({\cal A}\) | \({\cal B}\) | \(({\cal A}\lor {\cal B})\) |
|---|---|---|
| T | T | T |
| T | F | T |
| F | T | T |
| F | F | T |
Example
In the model where \(P\) is true, \(Q\) is false, \(R\) is true:
Propositions of the form \(\lnot {\cal A}\) (pronounced “not \({\cal A}\)”) are called negations.
Definition
A WFF of the form \(\lnot {\cal A}\) is true in a model when \({\cal A}\) is false in that model:
| \({\cal A}\) | \(\lnot {\cal A}\) |
|---|---|
| T | F |
| F | T |
Example
In the model where \(P\) is true, \(Q\) is false, \(R\) is true:
Propositions of the form \(({\cal A}\to {\cal B})\) (pronounced “\({\cal A}\) implies \({\cal B}\)”) are called implications.
In an implication \(({\cal A}\to {\cal B})\), we call \({\cal A}\) the premise and \({\cal B}\) the conclusion. (Others call \({\cal A}\) the antecedent and \({\cal B}\) the consequent.)
Definition
An implication of the form \(({\cal A}\to {\cal B})\) is true in a model if \({\cal A}\) is false in the model or \({\cal B}\) is true in the model.
Equivalently, the implication \(({\cal A}\to {\cal B})\) is only false when \({\cal A}\) is true in the model and \({\cal B}\) is false in the model.
| \({\cal A}\) | \({\cal B}\) | \(({\cal A}\to {\cal B})\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
Example
In the model where \(P\) is true, \(Q\) is false, \(R\) is true:
Implication defined as in this truth table is officially called material implication, and is the if-then relationship used in math and logic. Unlike the use of “if…then…” in everyday speech, there is no need for a cause-effect relationship between the premise and conclusion.
Example
The following are considered logically true statements in the model corresponding to the real world:
Definition
\(\top\) (pronounced “top”) is the WFF that is true in every model.
\(\bot\) (pronounced “bottom”) is the WFF that is false in every model.
Example
In the model where \(P\) is true, \(Q\) is false, \(R\) is true:
We use \(\top\) and \(\bot\) when writing logical formulas (WFFs) as strings of symbols, but write T and F when speaking philosophically of meanings, models, and truth or falsity.
(You can skip this if you feel OK about the truth table for implication.)
Folks often find the truth table for implication \((A \rightarrow B)\) confusing, especially in the cases where \(A\) is false. This section presents a set of conceptual explanations for why the truth table for logical implication, written as \(A \to B\), has the particular form that it does. The goal is to connect the formal symbols of logic to intuitive reasoning about claims, counterexamples, and failure cases.
This first section develops the idea of implication in terms of how you would try to show that it is false.
Suppose I make the claim:
“\(A\) implies \(B\).”
In everyday reasoning, one natural way to challenge this claim is to ask:
“What would it take to prove you wrong?”
To show that \(A \to B\) is not true, I would need to find a situation where:
This is exactly the case where the promise of implication is violated. If \(A\) really implies \(B\), then whenever \(A\) holds, \(B\) should also hold. A case where \(A\) holds but \(B\) does not is a counterexample.
The condition “\(A\) is true and \(B\) is false” can be written symbolically as:
\[ A \land \lnot B \]
This formula represents the failure of implication. It describes exactly the situation that would refute the claim \(A \to B\).
Now consider the statement:
“It is not the case that \(A\) is true and \(B\) is false.”
In symbols, this is:
\[ \lnot (A \land \lnot B) \]
This means there is no counterexample. There is no situation where \(A\) holds and \(B\) fails.
A central result in propositional logic is that:
\[ A \to B \;\equiv\; \lnot (A \land \lnot B) \]
These two expressions have the same truth table. They are true and false in exactly the same cases.
You should try working out the truth table for \(\lnot (A \land \lnot B)\) on your own and compare!
So instead of thinking of implication as a mysterious new operator, you can think of it as the negation of a very specific kind of failure.
Using this perspective, the truth table for \(A \to B\) follows directly from asking a single question in each row:
“Is this row a counterexample, where \(A\) is true and \(B\) is false?”
This explains why the only false case for implication is when \(A\) is true and \(B\) is false, and why all other combinations make \(A \to B\) come out true.
The first explanation focused on how implication fails. This one takes a more direct approach by asking:
“In what situations should \(A \to B\) count as true?”
We will build the meaning of implication by listing the cases where it is allowed to hold, and then combining those cases with a logical “or.”
If \(A\) is true and we want \(A \to B\) to hold, then \(B\) must also be true. Otherwise, we would have a violation of the implication.
This gives us our first allowed case:
\[ A \land B \]
This captures the situation where the rule is “used” and it works correctly.
Now consider what happens when \(A\) is false. If implication is like a rule that only places a requirement on \(B\) when \(A\) holds, then when \(A\) does not hold, the rule never gets triggered.
In this situation, \(B\) can be either true or false. Both possibilities are acceptable, because the condition that would force \(B\) to be true never arises.
We can write this as two subcases:
\[ \lnot A \land B \]
and
\[ \lnot A \land \lnot B \]
These two together can be combined into a single expression:
\[ \lnot A \land (B \lor \lnot B) \]
Since \(B \lor \lnot B\) is always true, this simplifies to:
\[ \lnot A \]
This reflects the idea that whenever \(A\) is false, the implication automatically holds, no matter what \(B\) is.
We now take the disjunction of all the allowed cases:
Written as a single formula, this is:
\[ (A \land B) \lor (\lnot A \land B) \lor (\lnot A \land \lnot B) \]
Using the standard rules of propositional logic, this expression simplifies to:
\[ \lnot A \lor B \]
A well-known logical equivalence is:
\[ A \to B \;\equiv\; \lnot A \lor B \]
So by listing all the situations in which the implication is allowed to be true, and then combining them with “or,” we arrive at a formula that has exactly the same truth table as \(A \to B\).
This gives a second way to understand implication: it is true either because \(A\) never happens, or because \(A\) happens and \(B\) successfully follows.
This explanation uses a familiar kind of mathematical statement, inequalities, to build intuition for why a false statement can imply anything.
Consider the statement:
\[ 5 \leq x \leq 10 \]
This says that \(x\) is between 5 and 10, inclusive.
Now consider a more specific statement:
\[ 6 \leq x \leq 8 \]
Every value of \(x\) that satisfies \(6 \leq x \leq 8\) also satisfies \(5 \leq x \leq 10\). So we can say:
\[ (6 \leq x \leq 8) \to (5 \leq x \leq 10) \]
The second statement implies the first because it is more restrictive. It allows fewer possible values of \(x\).
Now shrink the interval further:
\[ 7 \leq x \leq 8 \]
This still implies:
\[ 5 \leq x \leq 10 \]
Shrink again:
\[ 7 \leq x \leq 7 \]
This describes exactly one value of \(x\), namely \(x = 7\). This also implies that \(x\) is between 5 and 10.
Each time we shrink the interval, we reduce the number of values of \(x\) that satisfy the left-hand statement. The set of possible values gets smaller and smaller, but every remaining value still lies inside the larger interval from 5 to 10.
Now shrink one more time:
\[ 7 \leq x \leq 6 \]
There are no values of \(x\) that satisfy this. This statement is simply false.
Even in this case, the implication
\[ (7 \leq x \leq 6) \to (5 \leq x \leq 10) \]
is still considered true, because there is no value of \(x\) that makes the left-hand side true while making the right-hand side false.
You can think of the left-hand statement as the antecedent and the right-hand statement as the consequence.
As we keep shrinking the set of values that satisfy the antecedent, we reduce the chances for the implication to fail. Eventually, when the antecedent becomes impossible and has no satisfying values at all, there is no way to produce a counterexample.
This mirrors the logical rule for implication:
This interval example gives a concrete way to see why a false statement can imply anything. The more constrained the antecedent becomes, the fewer opportunities there are for the implication to break.
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