Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/dup04)

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

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

and S is the following TRS.

b(b(x1)) → b(b(b(b(b(b(x1)))))) (3)
a(a(x1)) → a(a(a(a(a(a(x1)))))) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(a(b(b(a(a(x1)))))) → a(a(b(b(b(b(a(a(x1)))))))) (1)
b(b(a(a(b(b(x1)))))) → b(b(a(a(a(a(b(b(x1)))))))) (2)
b(b(x1)) → b(b(b(b(b(b(x1)))))) (3)
a(a(x1)) → a(a(a(a(a(a(x1)))))) (4)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
aa(ab(bb(ba(aa(aa(x1)))))) → aa(ab(bb(bb(bb(ba(aa(aa(x1)))))))) (5)
aa(ab(bb(ba(aa(ab(x1)))))) → aa(ab(bb(bb(bb(ba(aa(ab(x1)))))))) (6)
bb(ba(aa(ab(bb(ba(x1)))))) → bb(ba(aa(aa(aa(ab(bb(ba(x1)))))))) (7)
bb(ba(aa(ab(bb(bb(x1)))))) → bb(ba(aa(aa(aa(ab(bb(bb(x1)))))))) (8)
bb(ba(x1)) → bb(bb(bb(bb(bb(ba(x1)))))) (9)
bb(bb(x1)) → bb(bb(bb(bb(bb(bb(x1)))))) (10)
aa(aa(x1)) → aa(aa(aa(aa(aa(aa(x1)))))) (11)
aa(ab(x1)) → aa(aa(aa(aa(aa(ab(x1)))))) (12)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the integers
[aa(x1)] =
0
0
0
+
1 0 1
0 1 0
0 0 0
· x1
[ab(x1)] =
0
1
0
+
1 0 0
0 0 0
0 1 0
· x1
[bb(x1)] =
0
0
0
+
1 0 0
0 0 1
0 0 0
· x1
[ba(x1)] =
0
0
0
+
1 0 0
0 0 0
0 1 0
· x1
all of the following rules can be deleted.
aa(ab(bb(ba(aa(ab(x1)))))) → aa(ab(bb(bb(bb(ba(aa(ab(x1)))))))) (6)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the integers
[aa(x1)] =
0
1
1
+
1 0 0
0 0 0
0 0 0
· x1
[ab(x1)] =
0
0
0
+
1 1 0
0 0 0
0 0 0
· x1
[bb(x1)] =
0
0
0
+
1 0 0
0 0 1
0 0 0
· x1
[ba(x1)] =
0
0
0
+
1 0 0
0 0 0
0 1 1
· x1
all of the following rules can be deleted.
aa(ab(bb(ba(aa(aa(x1)))))) → aa(ab(bb(bb(bb(ba(aa(aa(x1)))))))) (5)

1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the integers
[bb(x1)] =
0
0
0
+
1 0 0
0 0 0
1 0 0
· x1
[ba(x1)] =
0
0
0
+
1 0 1
0 0 0
0 0 0
· x1
[aa(x1)] =
0
0
1
+
1 0 0
0 0 0
0 1 0
· x1
[ab(x1)] =
0
1
0
+
1 0 1
0 0 0
0 0 0
· x1
all of the following rules can be deleted.
bb(ba(aa(ab(bb(ba(x1)))))) → bb(ba(aa(aa(aa(ab(bb(ba(x1)))))))) (7)
bb(ba(aa(ab(bb(bb(x1)))))) → bb(ba(aa(aa(aa(ab(bb(bb(x1)))))))) (8)

1.1.1.1.1.1 R is empty

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