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
- 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
- A => (B => A) [ND and SEQ]
- (A => (B => C)) => ((A => B) => (A => C)) [ND and SEQ]
- (A | (B & C)) => ((A | B) & (A | C)) [ND]
- ((A | B) & (A | C)) => (A | (B & C)) [SEQ]
- A => ~~A [ND]
- ~~A => A [SEQ]
- ~~~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.
- Give introduction and elimination rules for
equivalence without recourse to any other logical connectives.
- Display the local reductions that show the
local soundness of the elimination rules.
- Display the local expansion that show the
local completeness of the elimination rules.
- Prove that the two ways of defining logical
equivalence, as a notational definition or via introduction and
elimination rules, lead to the same theorems.
- Give the right and left rules for equivalence
in the sequent calculus that correspond to the introduction and
elimination rules in Question 2.1.
- 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.
- 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
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.