Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-67)

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

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

and S is the following TRS.

a(b(b(x1))) → b(c(b(x1))) (3)
c(c(b(x1))) → b(a(b(x1))) (4)
b(b(c(x1))) → c(c(a(x1))) (5)
b(a(a(x1))) → a(c(a(x1))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the integers
[b(x1)] =
0
0
0
+
2 1 0
2 1 0
2 1 0
· x1
[a(x1)] =
0
0
1
+
3 0 0
2 0 1
3 0 0
· x1
[c(x1)] =
0
0
0
+
3 0 0
3 0 0
3 0 0
· x1
all of the following rules can be deleted.
b(a(a(x1))) → c(a(c(x1))) (1)
b(a(a(x1))) → a(c(a(x1))) (6)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[a(x1)] =
0
0
+
1 1
1 1
· x1
[c(x1)] =
0
1
+
1 1
2 0
· x1
[b(x1)] =
1
0
+
2 0
2 0
· x1
all of the following rules can be deleted.
a(b(b(x1))) → b(c(b(x1))) (3)
c(c(b(x1))) → b(a(b(x1))) (4)
b(b(c(x1))) → c(c(a(x1))) (5)

1.1.1 Bounds

The given TRS is match-(raise)-bounded by 1. This is shown by the following automaton.