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 ttt2 @ 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#(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#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) (4)

1.1 Subterm Criterion Processor

We use the projection to multisets
π(p#) = { 2, 2 }
π(p) = { 2, 2 }
to remove the pairs:
p#(a(a(x0)),p(x1,p(a(x2),x3))) → 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)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[p(x1, x2)] =
0 0 0
0 0 0
0 0 1
· x1 +
0 0 1
0 0 0
1 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[p#(x1, x2)] =
1 0 1
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
0 1 0
0 0 0
1 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[b(x1)] =
0 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
together with the usable rule
p(a(a(x0)),p(x1,p(a(x2),x3))) → p(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pair
p#(a(a(x0)),p(x1,p(a(x2),x3))) → p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3))) (4)
could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.