Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/2-matchbox)

The rewrite relation of the following TRS is considered.

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

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(b(a(x1)))) b#(b(b(b(b(b(b(x1))))))) (4)
b#(b(b(a(x1)))) b#(b(b(x1))) (5)
c#(c(x1)) b#(b(x1)) (6)
b#(b(c(c(x1)))) c#(c(c(a(a(a(a(x1))))))) (7)
b#(b(c(c(x1)))) c#(a(a(a(a(x1))))) (8)
c#(c(x1)) b#(b(b(x1))) (9)
c#(c(x1)) b#(x1) (10)
b#(b(b(a(x1)))) b#(b(x1)) (11)
b#(b(c(c(x1)))) c#(c(a(a(a(a(x1)))))) (12)
b#(b(b(a(x1)))) b#(b(b(b(x1)))) (13)
b#(b(b(a(x1)))) b#(b(b(b(b(b(x1)))))) (14)
b#(b(b(a(x1)))) b#(b(b(b(b(b(b(b(x1)))))))) (15)
b#(b(b(a(x1)))) b#(x1) (16)
b#(b(b(a(x1)))) b#(b(b(b(b(x1))))) (17)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.