Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/torpa3)

The rewrite relation of the following TRS is considered.

b(b(x1)) → c(d(x1)) (1)
c(c(x1)) → d(d(d(x1))) (2)
c(x1) → g(x1) (3)
d(d(x1)) → c(f(x1)) (4)
d(d(d(x1))) → g(c(x1)) (5)
f(x1) → a(g(x1)) (6)
g(x1) → d(a(b(x1))) (7)
g(g(x1)) → b(c(x1)) (8)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

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

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[d(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[g(x1)] =
1 0 1
0 0 1
1 1 0
· x1 +
0 0 0
1 0 0
0 0 0
[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
0 0 0
1 0 0
[c(x1)] =
1 1 0
0 1 1
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
c(c(x1)) → d(d(d(x1))) (2)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(a) = 1 weight(a) = 1
prec(f) = 4 weight(f) = 5
prec(g) = 3 weight(g) = 4
prec(c) = 0 weight(c) = 1
prec(d) = 2 weight(d) = 3
prec(b) = 5 weight(b) = 0
all of the following rules can be deleted.
d(d(x1)) → c(f(x1)) (4)
f(x1) → a(g(x1)) (6)
g(x1) → d(a(b(x1))) (7)
g(g(x1)) → b(c(x1)) (8)

1.1.1.1.1 R is empty

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