Certification Problem

Input (TPDB SRS_Standard/Waldmann_06_SRS/jw5)

The rewrite relation of the following TRS is considered.

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

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.
b#(a(b(b(x1)))) b#(a(b(x1))) (5)
b#(a(b(b(x1)))) b#(b(a(b(x1)))) (6)
b#(a(b(b(x1)))) b#(b(b(a(b(x1))))) (7)
b#(a(a(b(b(x1))))) b#(a(a(b(x1)))) (8)
b#(a(a(b(b(x1))))) b#(b(a(a(b(x1))))) (9)
b#(a(a(b(b(x1))))) b#(a(b(b(a(a(b(x1))))))) (10)
b#(a(a(a(b(b(x1)))))) b#(a(a(a(b(x1))))) (11)
b#(a(a(a(b(b(x1)))))) b#(b(a(a(a(b(x1)))))) (12)
b#(a(a(a(b(b(x1)))))) b#(a(a(b(b(a(a(a(b(x1))))))))) (13)
b#(a(a(a(a(b(b(x1))))))) b#(a(a(a(a(b(x1)))))) (14)
b#(a(a(a(a(b(b(x1))))))) b#(b(a(a(a(a(b(x1))))))) (15)
b#(a(a(a(a(b(b(x1))))))) b#(a(a(a(b(b(a(a(a(a(b(x1))))))))))) (16)

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.