Certification Problem

Input (TPDB SRS_Standard/Wenzel_16/abcbaaaa-aaaaabcbaaabcb.srs)

The rewrite relation of the following TRS is considered.

a(b(c(b(a(a(a(a(x1)))))))) a(a(a(a(a(b(c(b(a(a(a(b(c(b(x1)))))))))))))) (1)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
ab(bc(cb(ba(aa(aa(aa(aa(x1)))))))) aa(aa(aa(aa(ab(bc(cb(ba(aa(aa(ab(bc(cb(ba(x1)))))))))))))) (2)
ab(bc(cb(ba(aa(aa(aa(ab(x1)))))))) aa(aa(aa(aa(ab(bc(cb(ba(aa(aa(ab(bc(cb(bb(x1)))))))))))))) (3)
ab(bc(cb(ba(aa(aa(aa(ac(x1)))))))) aa(aa(aa(aa(ab(bc(cb(ba(aa(aa(ab(bc(cb(bc(x1)))))))))))))) (4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
ab#(bc(cb(ba(aa(aa(aa(aa(x1)))))))) ab#(bc(cb(ba(x1)))) (5)
ab#(bc(cb(ba(aa(aa(aa(aa(x1)))))))) ab#(bc(cb(ba(aa(aa(ab(bc(cb(ba(x1)))))))))) (6)
ab#(bc(cb(ba(aa(aa(aa(ab(x1)))))))) ab#(bc(cb(bb(x1)))) (7)
ab#(bc(cb(ba(aa(aa(aa(ab(x1)))))))) ab#(bc(cb(ba(aa(aa(ab(bc(cb(bb(x1)))))))))) (8)
ab#(bc(cb(ba(aa(aa(aa(ac(x1)))))))) ab#(bc(cb(bc(x1)))) (9)
ab#(bc(cb(ba(aa(aa(aa(ac(x1)))))))) ab#(bc(cb(ba(aa(aa(ab(bc(cb(bc(x1)))))))))) (10)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.