Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-114)

The rewrite relation of the following TRS is considered.

b(a(a(a(x1)))) b(b(a(b(x1)))) (1)
a(b(a(b(x1)))) b(a(b(b(x1)))) (2)
a(a(b(a(x1)))) a(b(a(b(x1)))) (3)
b(b(b(b(x1)))) a(b(b(b(x1)))) (4)

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#(a(a(a(x1)))) b#(b(a(b(x1)))) (5)
b#(a(a(a(x1)))) b#(a(b(x1))) (6)
b#(a(a(a(x1)))) a#(b(x1)) (7)
b#(a(a(a(x1)))) b#(x1) (8)
a#(b(a(b(x1)))) b#(a(b(b(x1)))) (9)
a#(b(a(b(x1)))) a#(b(b(x1))) (10)
a#(b(a(b(x1)))) b#(b(x1)) (11)
a#(a(b(a(x1)))) a#(b(a(b(x1)))) (12)
a#(a(b(a(x1)))) b#(a(b(x1))) (13)
a#(a(b(a(x1)))) a#(b(x1)) (14)
a#(a(b(a(x1)))) b#(x1) (15)
b#(b(b(b(x1)))) a#(b(b(b(x1)))) (16)

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[b#(x1)] = 1 + 1 · x1
[a(x1)] = 1 + 1 · x1
[b(x1)] = 1 + 1 · x1
[a#(x1)] = 1 + 1 · x1
the pairs
b#(a(a(a(x1)))) b#(a(b(x1))) (6)
b#(a(a(a(x1)))) a#(b(x1)) (7)
b#(a(a(a(x1)))) b#(x1) (8)
a#(b(a(b(x1)))) a#(b(b(x1))) (10)
a#(b(a(b(x1)))) b#(b(x1)) (11)
a#(a(b(a(x1)))) b#(a(b(x1))) (13)
a#(a(b(a(x1)))) a#(b(x1)) (14)
a#(a(b(a(x1)))) b#(x1) (15)
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
-∞
-∞
+
-∞ -∞ 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
0
-∞
+
0 1 1
0 1 0
-∞ 0 -∞
· x1
[b(x1)] =
0
-∞
0
+
-∞ 0 1
0 -∞ 0
-∞ 0 1
· x1
[a#(x1)] =
0
-∞
-∞
+
0 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
the pair
a#(b(a(b(x1)))) b#(a(b(b(x1)))) (9)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.