Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-360)

The rewrite relation of the following TRS is considered.

a(x1) → b(x1) (1)
a(b(b(x1))) → c(x1) (2)
c(c(x1)) → a(b(c(a(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#(b(b(x1))) → c#(x1) (4)
c#(c(x1)) → a#(x1) (5)
c#(c(x1)) → c#(a(x1)) (6)
c#(c(x1)) → a#(b(c(a(x1)))) (7)

1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the arctic semiring over the integers
[c#(x1)] = 2 · x1 + -∞
[b(x1)] = 1 · x1 + -∞
[a#(x1)] = 0 · x1 + -∞
[a(x1)] = 1 · x1 + -∞
[c(x1)] = 3 · x1 + -∞
together with the usable rules
a(x1) → b(x1) (1)
a(b(b(x1))) → c(x1) (2)
c(c(x1)) → a(b(c(a(x1)))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(c(x1)) → a#(x1) (5)
c#(c(x1)) → c#(a(x1)) (6)
could be deleted.

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c#(x1)] =
0 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
0 0 1
1 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a#(x1)] =
0 1 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[a(x1)] =
0 0 1
1 1 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[c(x1)] =
0 0 0
0 0 1
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
together with the usable rules
a(x1) → b(x1) (1)
a(b(b(x1))) → c(x1) (2)
c(c(x1)) → a(b(c(a(x1)))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pair
a#(b(b(x1))) → c#(x1) (4)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.