Sequential composition of reductions.
Sequential composition of reductions.
This class is not intended to be used directly, but via the Reduction#|> operator.
A reduction without back-translation.
Represents a reduction of a problem together with a back-translation of the solutions.
Represents a reduction of a problem together with a back-translation of the solutions.
A problem P1 is reduced to a problem P2, a solution S2 to the problem P2 can then be translated back to a solution S1 of the problem P1.
A reduction that does not change the type of the problem.
Reduces finding an expansion proof for a sequent to finding a resolution proof of a clause set.
Reduces finding an LK proof for a sequent to finding a resolution proof of a clause set.
Reduces finding a resolution proof of a many-sorted clause set to the first-order case.
Reduces finding a resolution proof of a many-sorted clause set to the first-order case.
Sorts are simply ignored and we make a best effort to convert the resolution refutation back.
Reduces finding an expansion proof of a many-sorted sequent to the first-order case.
Reduces finding an expansion proof of a many-sorted sequent to the first-order case.
Sorts are simply ignored and we make a best effort to convert the expansion tree.
Simplifies the problem by grounding free variables.
Replaces the use of higher-order functions by fresh function symbols, together with axioms that axiomatize them.
Replaces lambda abstractions by fresh function symbols, together with axioms that axiomatize them.
Simplifies the problem of finding a resolution refutation of a many-sorted clause set by adding predicates for each of the sorts.
Simplifies the problem of finding a resolution refutation of a many-sorted clause set by adding predicates for each of the sorts. The resulting problem is still many-sorted.
Simplifies the problem of finding an expansion proof of a many-sorted sequent by adding predicates for each of the sorts.
Simplifies the problem of finding an expansion proof of a many-sorted sequent by adding predicates for each of the sorts. The resulting problem is still many-sorted.