To each atomic proposition P assign a natural number p. If that natural number is even, then its corresponding proposition is true. If the number is odd then the proposition is false.
If you have two numbers n and m and you multiply them, the result is even so long as either n or m is even. So multiplication is like or; P∨Q corresponds to pq.
Negation is like adding 1: even becomes odd and odd becomes even. So ¬P corresponds to p+1.
Consider addition: p+q is even if p and q are both even or both odd. So p+q is like the biconditional ↔.
Other connectives can be formed out of these. Take P∧Q: P∧Q is equivalent to ¬(¬P∨¬Q), which is (p+1)(q+1) + 1. So P∧Q corresponds to pq+p+q+2. P→Q is ¬P∨Q, which is (p+1)q.
The logical constants ⊤ and ⊥ correspond to nume…
To each atomic proposition P assign a natural number p. If that natural number is even, then its corresponding proposition is true. If the number is odd then the proposition is false.
If you have two numbers n and m and you multiply them, the result is even so long as either n or m is even. So multiplication is like or; P∨Q corresponds to pq.
Negation is like adding 1: even becomes odd and odd becomes even. So ¬P corresponds to p+1.
Consider addition: p+q is even if p and q are both even or both odd. So p+q is like the biconditional ↔.
Other connectives can be formed out of these. Take P∧Q: P∧Q is equivalent to ¬(¬P∨¬Q), which is (p+1)(q+1) + 1. So P∧Q corresponds to pq+p+q+2. P→Q is ¬P∨Q, which is (p+1)q.
The logical constants ⊤ and ⊥ correspond to numerical constants: ⊤ can be assigned 0 and ⊥ assigned 1, for instance.
Tautologies translate to algebraic expressions which are always even. For instance, P→P translates to (p+1)p, which is always even. P→(Q→P) translates to (p+1)(q+1)p, which is also always even. P∨¬P translates to p(p+1), always even.
Contradictions translate to algebraic expressions which are always odd. P∧¬P translates to (p+1)(p+2) + 1, which is always odd. And so on.
Inference rules can be understood as saying: if all the premises are even, then the conclusion will be even. Take conjunction elimination: P∧Q ⊨ P. This says that if (p+1)(q+1) + 1 is even then p is even, which ends up being right if you work it out. Modus ponens: P, P→Q ⊨ Q. This says that if p and (p+1)q are both even, then q is even. Again works out!
You can also work the other way: For any number p, p(p+1) is even. Translating into propositional logic, this says that P∨¬P is a tautology. We’ve proved the law of the excluded middle! It’s interesting to note that earlier we saw that P→P translates to (p+1)p. So in some sense the law of the excluded middle is just self-implication but with the two products reversed!