Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex25_Luc06_FR)

The rewrite relation of the following TRS is considered.

f(f(X)) → c(n__f(n__g(n__f(X)))) (1)
c(X) → d(activate(X)) (2)
h(X) → c(n__d(X)) (3)
f(X) → n__f(X) (4)
g(X) → n__g(X) (5)
d(X) → n__d(X) (6)
activate(n__f(X)) → f(activate(X)) (7)
activate(n__g(X)) → g(X) (8)
activate(n__d(X)) → d(X) (9)
activate(X) → X (10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
f(f(X)) → n__f(n__g(n__f(c(X)))) (11)
c(X) → activate(d(X)) (12)
h(X) → n__d(c(X)) (13)
f(X) → n__f(X) (4)
g(X) → n__g(X) (5)
d(X) → n__d(X) (6)
n__f(activate(X)) → activate(f(X)) (14)
n__g(activate(X)) → g(X) (15)
n__d(activate(X)) → d(X) (16)
activate(X) → X (10)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[activate(x1)] =
1 0 1
0 1 1
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[g(x1)] =
1 1 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[n__f(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 0 0
0 1 1
1 1 0
· x1 +
0 0 0
1 0 0
1 0 0
[d(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[n__d(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[h(x1)] =
1 1 0
1 1 0
0 0 0
· x1 +
1 0 0
1 0 0
1 0 0
[n__g(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
h(X) → n__d(c(X)) (13)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[activate(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[g(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[n__f(x1)] =
1 1 0
0 1 1
0 1 1
· x1 +
0 0 0
1 0 0
0 0 0
[c(x1)] =
1 1 0
0 1 1
0 1 1
· x1 +
0 0 0
1 0 0
0 0 0
[d(x1)] =
1 1 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 1 0
0 1 1
0 1 1
· x1 +
0 0 0
1 0 0
1 0 0
[n__d(x1)] =
1 1 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[n__g(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
n__f(activate(X)) → activate(f(X)) (14)
n__d(activate(X)) → d(X) (16)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(g) = 0 weight(g) = 4
prec(n__d) = 6 weight(n__d) = 1
prec(d) = 1 weight(d) = 2
prec(activate) = 3 weight(activate) = 1
prec(c) = 2 weight(c) = 4
prec(n__g) = 4 weight(n__g) = 3
prec(n__f) = 7 weight(n__f) = 0
prec(f) = 5 weight(f) = 4
all of the following rules can be deleted.
f(f(X)) → n__f(n__g(n__f(c(X)))) (11)
c(X) → activate(d(X)) (12)
f(X) → n__f(X) (4)
g(X) → n__g(X) (5)
d(X) → n__d(X) (6)
n__g(activate(X)) → g(X) (15)
activate(X) → X (10)

1.1.1.1.1 R is empty

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