Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-549)

The rewrite relation of the following TRS is considered.

a(b(x1)) → x1 (1)
a(c(x1)) → b(c(a(a(x1)))) (2)
c(b(x1)) → a(c(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)) → a#(a(x1)) (5)
a#(c(x1)) → c#(a(a(x1))) (6)
c#(b(x1)) → c#(x1) (7)
c#(b(x1)) → a#(c(x1)) (8)

1.1 Subterm Criterion Processor

We use the projection to multisets
π(c#) = { 1, 1 }
π(a#) = { 1 }
π(c) = { 1, 1 }
π(a) = { 1 }
π(b) = { 1 }
to remove the pairs:
a#(c(x1)) → a#(x1) (4)
a#(c(x1)) → a#(a(x1)) (5)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[c#(x1)] =
0 1
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[a(x1)] =
-∞ 0
0 -∞
· x1 +
0 -∞
0 -∞
[a#(x1)] =
0 1
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[b(x1)] =
-∞ 0
0 1
· x1 +
0 -∞
2 -∞
[c(x1)] =
0 1
-∞ 0
· x1 +
2 -∞
0 -∞
together with the usable rules
a(b(x1)) → x1 (1)
a(c(x1)) → b(c(a(a(x1)))) (2)
c(b(x1)) → a(c(x1)) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(b(x1)) → c#(x1) (7)
c#(b(x1)) → a#(c(x1)) (8)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.