Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z111)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[d(x1)] = x1 +
37/6
[c(x1)] = x1 +
29/3
[b(x1)] = x1 +
15
[a(x1)] = x1 +
22
all of the following rules can be deleted.
b(b(x1)) c(c(c(x1))) (4)
c(c(x1)) d(d(d(x1))) (5)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(d(d(x1))) a#(x1) (7)
a#(x1) c#(d(x1)) (8)
a#(a(x1)) c#(x1) (9)
a#(a(x1)) c#(c(x1)) (10)
a#(a(x1)) c#(c(c(x1))) (11)
a#(a(x1)) b#(c(c(c(x1)))) (12)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[d(x1)] = x1 +
2
[c(x1)] = x1 +
2
[b(x1)] = x1 +
6
[a(x1)] = x1 +
6
[c#(x1)] = x1 +
3
[b#(x1)] = x1 +
0
[a#(x1)] = x1 +
6
together with the usable rules
a(a(x1)) b(c(c(c(x1)))) (1)
b(c(x1)) d(d(d(d(x1)))) (2)
a(x1) d(c(d(x1))) (3)
c(d(d(x1))) a(x1) (6)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(d(d(x1))) a#(x1) (7)
a#(x1) c#(d(x1)) (8)
a#(a(x1)) c#(x1) (9)
a#(a(x1)) c#(c(x1)) (10)
a#(a(x1)) c#(c(c(x1))) (11)
a#(a(x1)) b#(c(c(c(x1)))) (12)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.