Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-110)

The rewrite relation of the following TRS is considered.

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

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(b(x1))) → a#(c(b(x1))) (7)
b#(b(a(x1))) → b#(x1) (8)
b#(b(a(x1))) → a#(b(x1)) (9)
b#(b(a(x1))) → a#(a(b(x1))) (10)
b#(a(c(x1))) → b#(a(x1)) (11)
b#(a(c(x1))) → a#(x1) (12)
b#(a(c(x1))) → a#(b(a(x1))) (13)
a#(a(a(x1))) → c#(c(a(x1))) (14)
a#(a(a(x1))) → c#(b(x1)) (15)
a#(a(a(x1))) → c#(a(x1)) (16)
a#(a(a(x1))) → b#(x1) (17)
a#(a(a(x1))) → b#(c(b(x1))) (18)
a#(a(a(x1))) → b#(a(a(x1))) (19)

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
b(b(a(x1))) → a(a(b(x1))) (1)
a(a(a(x1))) → c(c(a(x1))) (2)
a(a(a(x1))) → b(c(b(x1))) (3)
b(a(c(x1))) → a(b(a(x1))) (4)
c(c(b(x1))) → a(c(b(x1))) (5)
a(a(a(x1))) → b(a(a(x1))) (6)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(b(a(x1))) → b#(x1) (8)
b#(b(a(x1))) → a#(b(x1)) (9)
b#(a(c(x1))) → b#(a(x1)) (11)
b#(a(c(x1))) → a#(x1) (12)
a#(a(a(x1))) → c#(b(x1)) (15)
a#(a(a(x1))) → c#(a(x1)) (16)
a#(a(a(x1))) → b#(x1) (17)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.