Certification Problem

Input (TPDB SRS_Relative/Waldmann_23/size-10-alpha-2-num-57)

The relative rewrite relation R/S is considered where R is the following TRS

b(a(b(x))) → b(a(x)) (1)

and S is the following TRS.

a(a(x)) → a(a(b(x))) (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
ba(ab(bb(x))) → ba(ab(x)) (3)
ba(ab(ba(x))) → ba(aa(x)) (4)
aa(ab(x)) → aa(ab(bb(x))) (5)
aa(aa(x)) → aa(ab(ba(x))) (6)

1.1 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the integers
[ba(x1)] =
1
1
1
+
1 1 1
0 0 0
0 1 1
· x1
[ab(x1)] =
1
0
0
+
1 0 0
0 1 1
0 0 0
· x1
[bb(x1)] =
0
0
0
+
1 0 0
0 0 0
0 1 1
· x1
[aa(x1)] =
1
1
1
+
1 0 1
0 0 0
0 1 1
· x1
all of the following rules can be deleted.
ba(ab(ba(x))) → ba(aa(x)) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[ba(x1)] = 1 · x1
[ab(x1)] = 1 · x1
[bb(x1)] = 1 · x1
[aa(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
aa(aa(x)) → aa(ab(ba(x))) (6)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[ba(x1)] =
0
0
+
2 1
0 2
· x1
[ab(x1)] =
0
0
+
1 0
0 2
· x1
[bb(x1)] =
0
2
+
1 0
2 2
· x1
[aa(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
ba(ab(bb(x))) → ba(ab(x)) (3)

1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.