# Lecture 38: Proof systems

• proof trees and natural deduction

See the list of inference rules for a list of inference rules.

## Natural deduction

At each step of an informal proof, it is important to know - what you are currently trying to prove (the goal) - what assumptions you have made at that point (the assumptions) - what the next logical step is (which leads to a new set of goals and assumptions)

We can a single step in a proof this way:

Here the turnstyle () separates the assumptions from the goal.

We can then apply further rules to the new goals to build up a tree if logical inferences ending at the statement to be proved.

A logic is a set of allowed rules of inference.

A proof tree is valid in a logic if each step in the tree matches an acceptable inference rule. A formula ϕ is provable in the logic if there exists a valid proof tree with ϕ at the root.

## Natural deduction rules

We discussed this set of inference rules. The "introduction" rules capture the process you use to prove a certain kind of statement, while the "elimination" rules explain how to use a fact you have already proved. For example, to prove φψ, you first assume φ, and then your new goal is to prove ψ. This is captured by the introduction rule.

Similarly, if you have already proved φψ, and you can prove φ, then you can use these facts to conclude ψ. This is captured by the elimination rule.

The "reductio ad absurdum" rule is worth noting. It says that if, under some set of assumptions, you can prove both φ and ¬φ, then you can conclude anything you like under that same set of assumptions. Think of this as the end of a proof by contradiction; once you have reached a contradiction, you have completed your proof.

Thinking of this rule in terms of truth tables (see next lecture on the relationship between truth tables and proof systems), recall that φ1, …, φnφ means that in all of the rows of the truth table satisfying φ1, …, φn, φ evaluates to true. Thus, if φ1, …, φnφ and φ1, …, φn⊨¬φ, then every row of the truth table that satisfies φ1, …, φn must also satisfy both φ and ¬φ. But this is impossible, so there must not be any rows that satisfy all of the assumptions. Thus any formula ψ evaluates to true in all of the rows of the truth table that satisfy φ1, …, φn, since there are no such rows.

## Example proofs

We proved ⊢((PQ)→R)→(P→(QR)) and ⊢¬(PQ)→(¬P∨¬Q); see here (tex).

Note: I use "" in the list of assumptions as shorthand for "copy all of the assumptions from the line below".