Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-105)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(c(c(x1))) → b#(x1) (6)
c#(c(c(x1))) → b#(b(x1)) (7)
c#(c(c(x1))) → b#(b(b(x1))) (8)
b#(c(a(x1))) → a#(c(a(x1))) (9)
b#(b(b(x1))) → a#(b(x1)) (10)
b#(b(b(x1))) → a#(a(b(x1))) (11)
b#(a(a(x1))) → c#(x1) (12)
b#(a(a(x1))) → b#(c(x1)) (13)
b#(a(a(x1))) → a#(b(c(x1))) (14)
a#(c(b(x1))) → c#(a(a(x1))) (15)
a#(c(b(x1))) → a#(x1) (16)
a#(c(b(x1))) → a#(a(x1)) (17)

1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c(x1)] = x1 +
1
[b(x1)] = x1 +
1
[a(x1)] = x1 +
1
[c#(x1)] = x1 +
0
[b#(x1)] = x1 +
0
[a#(x1)] = x1 +
0
together with the usable rules
c(c(c(x1))) → b(b(b(x1))) (1)
b(a(a(x1))) → a(b(c(x1))) (2)
b(b(b(x1))) → a(a(b(x1))) (3)
b(c(a(x1))) → a(c(a(x1))) (4)
a(c(b(x1))) → c(a(a(x1))) (5)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(c(c(x1))) → b#(x1) (6)
c#(c(c(x1))) → b#(b(x1)) (7)
b#(b(b(x1))) → a#(b(x1)) (10)
b#(a(a(x1))) → c#(x1) (12)
b#(a(a(x1))) → b#(c(x1)) (13)
a#(c(b(x1))) → a#(x1) (16)
a#(c(b(x1))) → a#(a(x1)) (17)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.