Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup17)

The rewrite relation of the following TRS is considered.

a(a(a(a(x1)))) b(b(x1)) (1)
b(b(a(a(x1)))) a(a(b(b(x1)))) (2)
b(b(b(b(c(c(x1)))))) c(c(a(a(x1)))) (3)
b(b(b(b(x1)))) a(a(a(a(a(a(x1)))))) (4)
c(c(a(a(x1)))) b(b(a(a(c(c(x1)))))) (5)

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(x1)))) a#(b(b(x1))) (6)
c#(c(a(a(x1)))) b#(a(a(c(c(x1))))) (7)
b#(b(b(b(c(c(x1)))))) a#(a(x1)) (8)
c#(c(a(a(x1)))) c#(c(x1)) (9)
b#(b(b(b(c(c(x1)))))) c#(c(a(a(x1)))) (10)
b#(b(b(b(c(c(x1)))))) c#(a(a(x1))) (11)
b#(b(b(b(c(c(x1)))))) a#(x1) (12)
b#(b(b(b(x1)))) a#(a(a(x1))) (13)
c#(c(a(a(x1)))) a#(a(c(c(x1)))) (14)
c#(c(a(a(x1)))) c#(x1) (15)
c#(c(a(a(x1)))) b#(b(a(a(c(c(x1)))))) (16)
b#(b(b(b(x1)))) a#(x1) (17)
b#(b(b(b(x1)))) a#(a(a(a(a(a(x1)))))) (18)
b#(b(b(b(x1)))) a#(a(a(a(x1)))) (19)
b#(b(b(b(x1)))) a#(a(x1)) (20)
a#(a(a(a(x1)))) b#(x1) (21)
b#(b(a(a(x1)))) b#(b(x1)) (22)
b#(b(a(a(x1)))) a#(a(b(b(x1)))) (23)
c#(c(a(a(x1)))) a#(c(c(x1))) (24)
a#(a(a(a(x1)))) b#(b(x1)) (25)
b#(b(a(a(x1)))) b#(x1) (26)
b#(b(b(b(x1)))) a#(a(a(a(a(x1))))) (27)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.