Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/pair3swap)

The rewrite relation of the following TRS is considered.

p(a(a(x0)),p(x1,p(a(x2),x3))) → p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) (1)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
p#(a(a(x0)),p(x1,p(a(x2),x3))) → p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) (2)
p#(a(a(x0)),p(x1,p(a(x2),x3))) → p#(a(a(b(x1))),p(a(a(x0)),x3)) (3)
p#(a(a(x0)),p(x1,p(a(x2),x3))) → p#(a(a(x0)),x3) (4)

1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[p(x1, x2)] = 1 + 2 · x1 + 1 · x2
[a(x1)] = 1 · x1
[b(x1)] = 1 · x1
[p#(x1, x2)] = 2 · x1 + 1 · x2
the pairs
p#(a(a(x0)),p(x1,p(a(x2),x3))) → p#(a(a(b(x1))),p(a(a(x0)),x3)) (3)
p#(a(a(x0)),p(x1,p(a(x2),x3))) → p#(a(a(x0)),x3) (4)
and no rules could be deleted.

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the rationals with delta = 3/32
[p#(x1, x2)] = 0 + 1/4 · x1 + 1/2 · x2
[a(x1)] = 1/4 + 4 · x1
[p(x1, x2)] = 0 + 1/4 · x1 + 1/2 · x2
[b(x1)] = 0 + 0 · x1
the pair
p#(a(a(x0)),p(x1,p(a(x2),x3))) → p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) (2)
could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.