Certification Problem

Input (TPDB SRS_Standard/Waldmann_06_SRS/sym-5)

The rewrite relation of the following TRS is considered.

a(a(a(x1))) → b(b(b(x1))) (1)
b(b(b(b(x1)))) → a(b(b(a(x1)))) (2)

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(a(x1))) → b#(b(b(x1))) (3)
a#(a(a(x1))) → b#(b(x1)) (4)
a#(a(a(x1))) → b#(x1) (5)
b#(b(b(b(x1)))) → a#(b(b(a(x1)))) (6)
b#(b(b(b(x1)))) → b#(b(a(x1))) (7)
b#(b(b(b(x1)))) → b#(a(x1)) (8)
b#(b(b(b(x1)))) → a#(x1) (9)

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 + 1 · x1
[a(x1)] = 1 + 1 · x1
[b#(x1)] = 1 + 1 · x1
[b(x1)] = 1 + 1 · x1
the pairs
a#(a(a(x1))) → b#(b(x1)) (4)
a#(a(a(x1))) → b#(x1) (5)
b#(b(b(b(x1)))) → b#(b(a(x1))) (7)
b#(b(b(b(x1)))) → b#(a(x1)) (8)
b#(b(b(b(x1)))) → a#(x1) (9)
could be deleted.

1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 5 with strict dimension 1 over the arctic semiring over the naturals
[a#(x1)] =
-∞
-∞
-∞
-∞
-∞
+
0 0 0 -∞ 0
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
· x1
[a(x1)] =
0
0
1
0
-∞
+
-∞ 0 -∞ -∞ 0
1 -∞ 1 -∞ 0
-∞ 1 0 0 1
0 0 -∞ -∞ 0
0 0 0 -∞ 0
· x1
[b#(x1)] =
1
-∞
-∞
-∞
-∞
+
0 -∞ -∞ 0 0
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
· x1
[b(x1)] =
0
0
-∞
-∞
-∞
+
-∞ 0 0 -∞ 0
0 -∞ -∞ -∞ 1
0 -∞ 0 0 1
1 0 0 -∞ -∞
0 -∞ -∞ 0 0
· x1
the pair
a#(a(a(x1))) → b#(b(b(x1))) (3)
could be deleted.

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[b#(x1)] = 1 + 1 · x1
[b(x1)] = 0
[a#(x1)] = 0
[a(x1)] = 1 · x1
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
b#(b(b(b(x1)))) → a#(b(b(a(x1)))) (6)
could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.