Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-361)

The rewrite relation of the following TRS is considered.

b(a(b(a(x1)))) b(b(a(a(x1)))) (1)
a(a(a(b(x1)))) a(b(a(a(x1)))) (2)
a(a(b(a(x1)))) b(b(a(a(x1)))) (3)
a(b(b(b(x1)))) b(a(b(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.
a#(a(a(b(x1)))) b#(a(a(x1))) (5)
a#(a(b(a(x1)))) b#(a(a(x1))) (6)
a#(b(b(b(x1)))) b#(a(x1)) (7)
b#(a(b(a(x1)))) b#(a(a(x1))) (8)
a#(b(b(b(x1)))) b#(a(b(a(x1)))) (9)
a#(b(b(b(x1)))) a#(b(a(x1))) (10)
a#(b(b(b(x1)))) a#(x1) (11)
a#(a(b(a(x1)))) a#(a(x1)) (12)
b#(a(b(a(x1)))) a#(a(x1)) (13)
a#(a(b(a(x1)))) b#(b(a(a(x1)))) (14)
a#(a(a(b(x1)))) a#(a(x1)) (15)
a#(a(a(b(x1)))) a#(b(a(a(x1)))) (16)
b#(a(b(a(x1)))) b#(b(a(a(x1)))) (17)
a#(a(a(b(x1)))) a#(x1) (18)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.