Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/2)

The rewrite relation of the following TRS is considered.

c(c(c(b(x)))) → a(1,b(c(x))) (1)
b(c(b(c(x)))) → a(0,a(1,x)) (2)
a(0,x) → c(c(x)) (3)
a(1,x) → c(b(x)) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[0] =
2 0
2 0
[c(x1)] =
1 1
1 0
· x1 +
1 0
2 0
[a(x1, x2)] =
1 1
0 1
· x1 +
2 1
1 1
· x2 +
0 0
1 0
[b(x1)] =
1 1
1 0
· x1 +
0 0
1 0
[1] =
1 0
1 0
all of the following rules can be deleted.
b(c(b(c(x)))) → a(0,a(1,x)) (2)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(0) = 5 weight(0) = 5
prec(a) = 0 weight(a) = 1
prec(1) = 6 weight(1) = 1
prec(c) = 1 weight(c) = 1
prec(b) = 7 weight(b) = 0
all of the following rules can be deleted.
c(c(c(b(x)))) → a(1,b(c(x))) (1)
a(0,x) → c(c(x)) (3)
a(1,x) → c(b(x)) (4)

1.1.1 R is empty

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