Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/pair2simple1)

The rewrite relation of the following TRS is considered.

p(a(x0),p(a(b(x1)),x2)) → p(a(b(a(x2))),p(a(a(x1)),x2)) (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(x0),p(a(b(x1)),x2)) → p#(a(a(x1)),x2) (2)
p#(a(x0),p(a(b(x1)),x2)) → p#(a(b(a(x2))),p(a(a(x1)),x2)) (3)

1.1 Subterm Criterion Processor

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

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[b(x1)] =
0 0
0 0
· x1 +
0 0
2 0
[p#(x1, x2)] =
0 0
0 0
· x1 +
2 0
0 0
· x2 +
0 0
0 0
[a(x1)] =
0 1
0 0
· x1 +
0 0
0 0
[p(x1, x2)] =
1 0
0 0
· x1 +
1 0
0 0
· x2 +
1 0
0 0
together with the usable rule
p(a(x0),p(a(b(x1)),x2)) → p(a(b(a(x2))),p(a(a(x1)),x2)) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pair
p#(a(x0),p(a(b(x1)),x2)) → p#(a(b(a(x2))),p(a(a(x1)),x2)) (3)
could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.