Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-33)

The rewrite relation of the following TRS is considered.

b(b(b(b(x1)))) → a(a(a(a(x1)))) (1)
a(b(a(a(x1)))) → b(b(b(b(x1)))) (2)
b(a(b(a(x1)))) → a(a(b(a(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.
b#(b(b(b(x1)))) → a#(a(a(a(x1)))) (4)
b#(b(b(b(x1)))) → a#(a(a(x1))) (5)
b#(b(b(b(x1)))) → a#(a(x1)) (6)
b#(b(b(b(x1)))) → a#(x1) (7)
a#(b(a(a(x1)))) → b#(b(b(b(x1)))) (8)
a#(b(a(a(x1)))) → b#(b(b(x1))) (9)
a#(b(a(a(x1)))) → b#(b(x1)) (10)
a#(b(a(a(x1)))) → b#(x1) (11)
b#(a(b(a(x1)))) → a#(a(b(a(x1)))) (12)

1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[b(x1)] = 2 + 1 · x1
[a(x1)] = 2 + 1 · x1
[b#(x1)] = 2 · x1
[a#(x1)] = 2 · x1
the pairs
b#(b(b(b(x1)))) → a#(a(a(x1))) (5)
b#(b(b(b(x1)))) → a#(a(x1)) (6)
b#(b(b(b(x1)))) → a#(x1) (7)
a#(b(a(a(x1)))) → b#(b(b(x1))) (9)
a#(b(a(a(x1)))) → b#(b(x1)) (10)
a#(b(a(a(x1)))) → b#(x1) (11)
and no rules 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
[b#(x1)] =
0
-∞
-∞
-∞
-∞
+
-∞ -∞ -∞ -∞ 0
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
· x1
[b(x1)] =
0
0
0
0
-∞
+
0 0 0 0 0
0 -∞ -∞ -∞ -∞
0 0 0 -∞ 0
-∞ 0 0 -∞ 1
-∞ 0 -∞ -∞ 0
· x1
[a#(x1)] =
0
-∞
-∞
-∞
-∞
+
-∞ -∞ 0 0 0
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
-∞ -∞ -∞ -∞ -∞
· x1
[a(x1)] =
0
0
0
0
0
+
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1
the pair
a#(b(a(a(x1)))) → b#(b(b(b(x1)))) (8)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.