Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-113)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

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

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 · x1
[a(x1)] = 1 + 1 · x1
[b(x1)] = 1 + 1 · x1
the pairs
a#(a(b(a(x1)))) a#(b(x1)) (5)
a#(b(a(a(x1)))) a#(a(a(x1))) (7)
could be deleted.

1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[a#(x1)] =
0
-∞
-∞
+
0 -∞ 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
-∞
-∞
+
-∞ 0 0
-∞ 0 0
0 0 0
· x1
[b(x1)] =
1
0
0
+
-∞ 1 0
-∞ -∞ -∞
-∞ 0 -∞
· x1
the pair
a#(b(a(a(x1)))) a#(a(a(a(x1)))) (6)
could be deleted.

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 · x1
[a(x1)] = 1
[b(x1)] = 0
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
a#(a(b(a(x1)))) a#(b(a(b(x1)))) (4)
could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.