Certification Problem

Input (TPDB SRS_Standard/Waldmann_06_SRS/jw1)

The rewrite relation of the following TRS is considered.

b(b(b(x1))) a(a(a(x1))) (1)
a(a(a(x1))) b(a(b(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.
b#(b(b(x1))) a#(a(a(x1))) (3)
b#(b(b(x1))) a#(a(x1)) (4)
b#(b(b(x1))) a#(x1) (5)
a#(a(a(x1))) b#(a(b(x1))) (6)
a#(a(a(x1))) a#(b(x1)) (7)
a#(a(a(x1))) b#(x1) (8)

1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 + 2 · x1
[a(x1)] = 1 + 2 · x1
[b#(x1)] = 2 + 2 · x1
[a#(x1)] = 2 + 2 · x1
the pairs
b#(b(b(x1))) a#(a(x1)) (4)
b#(b(b(x1))) a#(x1) (5)
a#(a(a(x1))) a#(b(x1)) (7)
a#(a(a(x1))) b#(x1) (8)
and no rules 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
[b#(x1)] =
-∞
-∞
-∞
+
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
0
0
0
+
1 0 0
0 -∞ -∞
0 0 -∞
· x1
[a#(x1)] =
0
-∞
-∞
+
-∞ 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
0
0
+
-∞ 0 0
0 0 -∞
0 1 -∞
· x1
the pair
b#(b(b(x1))) a#(a(a(x1))) (3)
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 + 1 · x1
[b#(x1)] = 1
[b(x1)] = 1 + 1 · x1
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
a#(a(a(x1))) b#(a(b(x1))) (6)
could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.