Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-142)

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

a(a(b(x1))) → a(c(b(x1))) (1)
c(c(c(x1))) → a(b(a(x1))) (2)
b(c(b(x1))) → b(a(c(x1))) (3)

and S is the following TRS.

b(a(a(x1))) → a(c(c(x1))) (4)
b(b(b(x1))) → a(b(b(x1))) (5)
b(b(c(x1))) → b(c(c(x1))) (6)
b(b(c(x1))) → c(a(b(x1))) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[a(x1)] =
0
0
+
2 0
0 0
· x1
[b(x1)] =
0
0
+
2 1
2 1
· x1
[c(x1)] =
0
1
+
2 0
0 2
· x1
all of the following rules can be deleted.
b(c(b(x1))) → b(a(c(x1))) (3)
b(b(c(x1))) → c(a(b(x1))) (7)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[a(x1)] =
0
0
+
2 0
1 0
· x1
[b(x1)] =
0
0
+
1 2
1 1
· x1
[c(x1)] =
0
2
+
2 0
1 0
· x1
all of the following rules can be deleted.
b(b(c(x1))) → b(c(c(x1))) (6)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[a(x1)] =
0
0
+
1 0
0 0
· x1
[b(x1)] =
0
2
+
1 2
0 0
· x1
[c(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
b(b(b(x1))) → a(b(b(x1))) (5)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[a(x1)] =
2
0
+
2 0
2 0
· x1
[b(x1)] =
0
0
+
1 1
1 1
· x1
[c(x1)] =
1
0
+
2 0
2 0
· x1
all of the following rules can be deleted.
a(a(b(x1))) → a(c(b(x1))) (1)
c(c(c(x1))) → a(b(a(x1))) (2)
b(a(a(x1))) → a(c(c(x1))) (4)

1.1.1.1.1 R is empty

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