Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-105)

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

c(c(c(x1))) → b(b(b(x1))) (1)

and S is the following TRS.

b(a(a(x1))) → a(b(c(x1))) (2)
b(b(b(x1))) → a(a(b(x1))) (3)
b(c(a(x1))) → a(c(a(x1))) (4)
a(c(b(x1))) → c(a(a(x1))) (5)

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
[b(x1)] =
1 0 1
0 1 1
0 1 0
· x1 +
0 0 0
0 0 0
1 0 0
[c(x1)] =
1 0 1
0 1 1
0 1 0
· x1 +
0 0 0
0 0 0
1 0 0
[a(x1)] =
1 0 0
0 1 1
0 1 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
b(b(b(x1))) → a(a(b(x1))) (3)
b(c(a(x1))) → a(c(a(x1))) (4)

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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