Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-563)

The rewrite relation of the following TRS is considered.

a(b(x1)) x1 (1)
a(c(x1)) c(c(b(a(x1)))) (2)
b(c(x1)) a(b(x1)) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(c(x1)) a#(x1) (4)
a#(c(x1)) b#(a(x1)) (5)
b#(c(x1)) b#(x1) (6)
b#(c(x1)) a#(b(x1)) (7)

1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the arctic semiring over the integers
[b#(x1)] = 0 · x1 + 0
[a(x1)] = 2 · x1 + 0
[a#(x1)] = 0 · x1 + 0
[b(x1)] = -2 · x1 + 0
[c(x1)] = 2 · x1 + 4
together with the usable rules
a(b(x1)) x1 (1)
a(c(x1)) c(c(b(a(x1)))) (2)
b(c(x1)) a(b(x1)) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(c(x1)) a#(x1) (4)
b#(c(x1)) b#(x1) (6)
b#(c(x1)) a#(b(x1)) (7)
could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.