Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/06)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
aa(ab(ba(aa(x1)))) → ab(bb(ba(x1))) (3)
aa(ab(ba(ab(x1)))) → ab(bb(bb(x1))) (4)
bb(ba(x1)) → ba(aa(aa(aa(x1)))) (5)
bb(bb(x1)) → ba(aa(aa(ab(x1)))) (6)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
aa#(ab(ba(aa(x1)))) → bb#(ba(x1)) (7)
aa#(ab(ba(ab(x1)))) → bb#(bb(x1)) (8)
aa#(ab(ba(ab(x1)))) → bb#(x1) (9)
bb#(ba(x1)) → aa#(aa(aa(x1))) (10)
bb#(ba(x1)) → aa#(aa(x1)) (11)
bb#(ba(x1)) → aa#(x1) (12)
bb#(bb(x1)) → aa#(aa(ab(x1))) (13)
bb#(bb(x1)) → aa#(ab(x1)) (14)

1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[aa#(x1)] =
0
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[ab(x1)] =
0
0
0
+
0 -∞ 0
0 -∞ -∞
0 0 -∞
· x1
[ba(x1)] =
0
1
0
+
-∞ -∞ 0
0 0 1
-∞ 0 0
· x1
[aa(x1)] =
0
0
-∞
+
0 0 0
0 0 0
-∞ -∞ 0
· x1
[bb#(x1)] =
0
-∞
-∞
+
0 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[bb(x1)] =
0
0
0
+
0 0 -∞
1 0 0
-∞ 0 0
· x1
the pair
aa#(ab(ba(ab(x1)))) → bb#(x1) (9)
could be deleted.

1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[aa#(x1)] =
0
-∞
-∞
+
-∞ 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[ab(x1)] =
0
0
0
+
0 0 1
0 0 0
0 1 0
· x1
[ba(x1)] =
0
-∞
0
+
0 0 0
0 0 0
0 0 0
· x1
[aa(x1)] =
-∞
-∞
-∞
+
0 0 -∞
0 -∞ -∞
0 0 0
· x1
[bb#(x1)] =
-∞
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[bb(x1)] =
0
0
0
+
-∞ 1 1
0 0 0
0 0 0
· x1
the pair
aa#(ab(ba(ab(x1)))) → bb#(bb(x1)) (8)
could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.