15-815 Automated Theorem Proving Assignment 1: Natural Deduction and Sequent Calculus

The assignment is worth 80 pts, due Thu Jan 29

• Reading: Chs. 2.1 and 3.3
• Code: none

Problem 1: Deduction (35 pts)

For each of the following propositions, give proofs in the indicated format if they exist, or state that no proof exists with the rules of intuitionistic logic. We use the following syntax

```A ::= P
| ~ A         % negation
| A & A       % conjunction
| A | A       % disjunction
| A => A      % implication
| (A)         % parentheses to override precedence
Operator Precedence  ~ > & > | > =>
& | are left associative
=> is right associative
```
1. A => (B => A) [ND and SEQ]
2. (A => (B => C)) => ((A => B) => (A => C)) [ND and SEQ]
3. (A | (B & C)) => ((A | B) & (A | C)) [ND]
4. ((A | B) & (A | C)) => (A | (B & C)) [SEQ]
5. A => ~~A [ND]
6. ~~A => A [SEQ]
7. ~~~A => ~A [SEQ]

Problem 2: Logical Equivalence (35 pts)

Logical equivalence, A <=> B is usually defined as (A => B) & (B => A). In this problem we explore the definition of equivalence using introduction and elimination rules.

1. Give introduction and elimination rules for equivalence without recourse to any other logical connectives.
2. Display the local reductions that show the local soundness of the elimination rules.
3. Display the local expansion that show the local completeness of the elimination rules.
4. Prove that the two ways of defining logical equivalence, as a notational definition or via introduction and elimination rules, lead to the same theorems.
5. Give the right and left rules for equivalence in the sequent calculus that correspond to the introduction and elimination rules in Question 2.1.
6. Propose at least one alternative set of introduction and/or elimination rules for logical equivalence which also satisfies the criteria of Questions 2.1 and 2.4.
7. Show the sequent calculus rules corresponding to the alternatives introduction and elimination rules from Question 2.6

Problem 3: Contraction (10 pts)

Show by means of a counterexample that the sequent calculus rule (using G for Gamma)

``` G ==> A  G, B ==> C
------------------- =>L'
G, A => B ==> C
```

is not complete (without a rule of contraction). In other words, give a theorem we can prove with the rule =>L given in lecture where A => B is retained in the left premise, but not in the system with =>L' instead.

fp@cs
Frank Pfenning