Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup05)

The rewrite relation of the following TRS is considered.

a(a(s(s(x1)))) s(s(a(a(x1)))) (1)
b(b(a(a(b(b(s(s(x1)))))))) a(a(b(b(s(s(a(a(x1)))))))) (2)
b(b(a(a(b(b(b(b(x1)))))))) a(a(b(b(a(a(b(b(x1)))))))) (3)
a(a(b(b(a(a(a(a(x1)))))))) b(b(a(a(b(b(a(a(x1)))))))) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(a(a(b(b(s(s(x1)))))))) a#(b(b(s(s(a(a(x1))))))) (5)
b#(b(a(a(b(b(s(s(x1)))))))) a#(x1) (6)
a#(a(s(s(x1)))) a#(a(x1)) (7)
b#(b(a(a(b(b(b(b(x1)))))))) b#(b(a(a(b(b(x1)))))) (8)
b#(b(a(a(b(b(b(b(x1)))))))) a#(a(b(b(x1)))) (9)
b#(b(a(a(b(b(b(b(x1)))))))) a#(b(b(x1))) (10)
a#(a(s(s(x1)))) a#(x1) (11)
a#(a(b(b(a(a(a(a(x1)))))))) b#(a(a(x1))) (12)
b#(b(a(a(b(b(b(b(x1)))))))) a#(a(b(b(a(a(b(b(x1)))))))) (13)
b#(b(a(a(b(b(b(b(x1)))))))) b#(a(a(b(b(x1))))) (14)
b#(b(a(a(b(b(s(s(x1)))))))) a#(a(x1)) (15)
a#(a(b(b(a(a(a(a(x1)))))))) a#(a(b(b(a(a(x1)))))) (16)
a#(a(b(b(a(a(a(a(x1)))))))) b#(b(a(a(x1)))) (17)
a#(a(b(b(a(a(a(a(x1)))))))) b#(a(a(b(b(a(a(x1))))))) (18)
b#(b(a(a(b(b(s(s(x1)))))))) b#(b(s(s(a(a(x1)))))) (19)
b#(b(a(a(b(b(s(s(x1)))))))) a#(a(b(b(s(s(a(a(x1)))))))) (20)
b#(b(a(a(b(b(b(b(x1)))))))) a#(b(b(a(a(b(b(x1))))))) (21)
a#(a(b(b(a(a(a(a(x1)))))))) b#(b(a(a(b(b(a(a(x1)))))))) (22)
b#(b(a(a(b(b(s(s(x1)))))))) b#(s(s(a(a(x1))))) (23)
a#(a(b(b(a(a(a(a(x1)))))))) a#(b(b(a(a(x1))))) (24)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.