Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-371)

The rewrite relation of the following TRS is considered.

a(a(a(a(x1)))) b(a(a(a(x1)))) (1)
b(b(a(b(x1)))) a(a(b(b(x1)))) (2)
a(a(b(a(x1)))) a(a(a(b(x1)))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(a(b(x1)))) b#(b(x1)) (4)
b#(b(a(b(x1)))) a#(b(b(x1))) (5)
b#(b(a(b(x1)))) a#(a(b(b(x1)))) (6)
a#(a(b(a(x1)))) b#(x1) (7)
a#(a(b(a(x1)))) a#(b(x1)) (8)
a#(a(b(a(x1)))) a#(a(b(x1))) (9)
a#(a(b(a(x1)))) a#(a(a(b(x1)))) (10)
a#(a(a(a(x1)))) b#(a(a(a(x1)))) (11)

1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[b(x1)] = x1 +
1
[a(x1)] = x1 +
1
[b#(x1)] = x1 +
0
[a#(x1)] = x1 +
0
together with the usable rules
a(a(a(a(x1)))) b(a(a(a(x1)))) (1)
b(b(a(b(x1)))) a(a(b(b(x1)))) (2)
a(a(b(a(x1)))) a(a(a(b(x1)))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(b(a(b(x1)))) b#(b(x1)) (4)
b#(b(a(b(x1)))) a#(b(b(x1))) (5)
a#(a(b(a(x1)))) b#(x1) (7)
a#(a(b(a(x1)))) a#(b(x1)) (8)
a#(a(b(a(x1)))) a#(a(b(x1))) (9)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.