Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-186)

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

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

and S is the following TRS.

a(b(c(x1))) → b(b(b(x1))) (4)
c(a(c(x1))) → c(b(a(x1))) (5)
b(c(b(x1))) → b(c(c(x1))) (6)
a(a(a(x1))) → a(c(c(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 3 with strict dimension 1 over the integers
[b(x1)] =
1
1
2
+
1 1 0
1 1 1
0 0 0
· x1
[c(x1)] =
3
0
0
+
2 0 0
0 2 1
0 0 0
· x1
[a(x1)] =
3
3
2
+
2 0 0
2 0 0
0 0 0
· x1
all of the following rules can be deleted.
b(c(b(x1))) → a(b(b(x1))) (1)
b(c(a(x1))) → a(b(a(x1))) (3)
a(b(c(x1))) → b(b(b(x1))) (4)
c(a(c(x1))) → c(b(a(x1))) (5)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a(x1)] = 1 · x1
[c(x1)] = 1 · x1
[b(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
b(c(b(x1))) → b(c(c(x1))) (6)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(a) = 1 weight(a) = 1
prec(c) = 0 weight(c) = 1
all of the following rules can be deleted.
a(c(c(x1))) → c(a(a(x1))) (2)
a(a(a(x1))) → a(c(c(x1))) (7)

1.1.1.1 R is empty

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