15-815 Automated Theorem Proving
Assignment 3: Invertibility
The assignment is worth 80 pts, due Tue Feb 17
Problem 1: Invertibility (30 pts)
An inference rule is invertible if the premises are
derivable whenever the conclusion is. In the classical sequent
calculus of Section 3.7 of the notes, all rules are invertible
simply because each sequent in the premises has more assumptions
(about truth and falsehood) than the conclusion. In this
exercise we are only concerned with the logical operators
of conjunction, disjunction, and negation.
- Give a formulation of the inference rules
that is contraction-free in the sense that the principal formula
of the conclusion does not reappear in any of the premises. Take
care to define your rules in such a way that they remain
- Prove that your formulation is sound
and complete with respect to the calculus given in the notes.
- Prove that your rules are indeed invertible.
Problem 2: Implementation (20 pts)
Consider the following signature given the library functions
from Assignment 2 (see file
signature CL =
val decide : P.Prop -> bool
- Define a structure C1_yourid :> CL
that implements your rules from Problem 1. Your implementation
should take advantage of the invertibility of the rules, but
may otherwise be as inefficient as you like.
- Define a structure C2_yourid :> CL
that works via translation to intuitionistic propositional
logic and calls your solution from Assignment 2.
- Discuss the possibility of giving a principled extension
of Dyckhoff's decision procedure so it is more efficient
on the formulas translated from classical logic.
Also submit a test suite TC_yourid (D : CL) :> TEST
(as in Assignment 2) that tests your
implementation on some classical propositions.
Problem 3: Strong Invertibility (30 pts)
An inference rule is strongly invertible if the premises are
derivable whenever the conclusion is and, moreover, the premises have
derivations of the same or equal height than the conclusion. Here we
define the height of a derivation as the number of inferences on the
longest path from the conclusion to an initial sequent or axiom.
- Prove that right rules for conjunction
and implication in the intuitionistic sequent calculus
are strongly invertible.
- Prove that a strict version
of the disjunction left rule in the intuitionistic sequent
calculus that erases the disjunction from both branches, is strongly
- Give a counterexample to show that we cannot
require the premises of a strongly invertible rule to have proofs
of strictly smaller height than the conclusion.