Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z104)

The rewrite relation of the following TRS is considered.

c(c(c(a(x1)))) d(d(x1)) (1)
d(b(x1)) c(c(x1)) (2)
c(x1) a(a(a(a(x1)))) (3)
d(x1) b(b(b(b(x1)))) (4)
b(d(x1)) c(c(x1)) (5)
a(c(c(c(x1)))) d(d(x1)) (6)

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.
d#(b(x1)) c#(x1) (7)
b#(d(x1)) c#(x1) (8)
d#(x1) b#(b(b(b(x1)))) (9)
c#(x1) a#(a(x1)) (10)
c#(c(c(a(x1)))) d#(d(x1)) (11)
c#(c(c(a(x1)))) d#(x1) (12)
d#(x1) b#(b(b(x1))) (13)
c#(x1) a#(a(a(a(x1)))) (14)
c#(x1) a#(x1) (15)
b#(d(x1)) c#(c(x1)) (16)
d#(x1) b#(x1) (17)
a#(c(c(c(x1)))) d#(d(x1)) (18)
d#(b(x1)) c#(c(x1)) (19)
c#(x1) a#(a(a(x1))) (20)
d#(x1) b#(b(x1)) (21)
a#(c(c(c(x1)))) d#(x1) (22)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.