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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[c(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
b(c(b(x1))) b(a(c(x1))) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 1 0
0 1 1
1 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[c(x1)] =
1 0 0
1 1 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
b(b(c(x1))) c(a(b(x1))) (4)
b(b(b(x1))) a(b(b(x1))) (5)
b(b(c(x1))) b(c(c(x1))) (6)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a(x1)] = 2 · x1 + 20
[b(x1)] = 2 · x1 + 6
[c(x1)] = 2 · x1 + 16
all of the following rules can be deleted.
a(a(b(x1))) a(c(b(x1))) (1)
b(a(a(x1))) a(c(c(x1))) (7)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(c) = 0 weight(c) = 5
prec(a) = 1 weight(a) = 1
prec(b) = 3 weight(b) = 5
all of the following rules can be deleted.
c(c(c(x1))) a(b(a(x1))) (2)

1.1.1.1.1 R is empty

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