Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-81)

The rewrite relation of the following TRS is considered.

a(b(b(x1))) → c(b(a(x1))) (1)
a(a(c(x1))) → b(c(a(x1))) (2)
a(a(b(x1))) → a(c(b(x1))) (3)
b(a(a(x1))) → c(a(a(x1))) (4)
c(b(b(x1))) → b(a(a(x1))) (5)
c(b(a(x1))) → b(c(c(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#(b(b(x1))) → b#(a(a(x1))) (7)
c#(b(b(x1))) → a#(x1) (8)
c#(b(b(x1))) → a#(a(x1)) (9)
c#(b(a(x1))) → c#(x1) (10)
c#(b(a(x1))) → c#(c(x1)) (11)
c#(b(a(x1))) → b#(c(c(x1))) (12)
b#(a(a(x1))) → c#(a(a(x1))) (13)
a#(b(b(x1))) → c#(b(a(x1))) (14)
a#(b(b(x1))) → b#(a(x1)) (15)
a#(b(b(x1))) → a#(x1) (16)
a#(a(c(x1))) → c#(a(x1)) (17)
a#(a(c(x1))) → b#(c(a(x1))) (18)
a#(a(c(x1))) → a#(x1) (19)
a#(a(b(x1))) → c#(b(x1)) (20)
a#(a(b(x1))) → a#(c(b(x1))) (21)

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 +
3
[b(x1)] = x1 +
3
[a(x1)] = x1 +
3
[c#(x1)] = x1 +
1
[b#(x1)] = x1 +
1
[a#(x1)] = x1 +
2
together with the usable rules
a(b(b(x1))) → c(b(a(x1))) (1)
a(a(c(x1))) → b(c(a(x1))) (2)
a(a(b(x1))) → a(c(b(x1))) (3)
b(a(a(x1))) → c(a(a(x1))) (4)
c(b(b(x1))) → b(a(a(x1))) (5)
c(b(a(x1))) → b(c(c(x1))) (6)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(b(b(x1))) → a#(x1) (8)
c#(b(b(x1))) → a#(a(x1)) (9)
c#(b(a(x1))) → c#(x1) (10)
c#(b(a(x1))) → c#(c(x1)) (11)
a#(b(b(x1))) → c#(b(a(x1))) (14)
a#(b(b(x1))) → b#(a(x1)) (15)
a#(b(b(x1))) → a#(x1) (16)
a#(a(c(x1))) → c#(a(x1)) (17)
a#(a(c(x1))) → b#(c(a(x1))) (18)
a#(a(c(x1))) → a#(x1) (19)
a#(a(b(x1))) → c#(b(x1)) (20)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.