Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex16_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(X,X) → a__f(a,b) (1)
a__b → a (2)
mark(f(X1,X2)) → a__f(mark(X1),X2) (3)
mark(b) → a__b (4)
mark(a) → a (5)
a__f(X1,X2) → f(X1,X2) (6)
a__b → b (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
[f(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 1
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[a] =
0 0 0
0 0 0
0 0 0
[a__b] =
1 0 0
0 0 0
1 0 0
[mark(x1)] =
1 0 0
1 0 0
1 0 0
· x1 +
1 0 0
0 0 0
1 0 0
[a__f(x1, x2)] =
1 0 0
0 0 0
0 1 0
· x1 +
1 0 1
1 0 0
1 0 1
· x2 +
0 0 0
0 0 0
1 0 0
[b] =
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
a__b → a (2)
mark(a) → a (5)
a__b → b (7)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1, x2)] =
1 0 1
0 1 0
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[a] =
0 0 0
0 0 0
0 0 0
[a__b] =
0 0 0
0 0 0
1 0 0
[mark(x1)] =
1 1 1
0 0 0
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
[a__f(x1, x2)] =
1 0 1
0 1 0
0 1 1
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[b] =
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
mark(b) → a__b (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1, x2)] =
1 0 1
0 0 0
0 0 1
· x1 +
1 1 0
0 0 0
1 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[a] =
0 0 0
0 0 0
0 0 0
[mark(x1)] =
1 0 1
0 0 1
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[a__f(x1, x2)] =
1 0 1
0 0 1
0 0 1
· x1 +
1 1 0
1 0 0
1 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[b] =
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
mark(f(X1,X2)) → a__f(mark(X1),X2) (3)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals
[f(x1, x2)] =
1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
· x1 +
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
· x2 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[a] =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[a__f(x1, x2)] =
1 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0
· x1 +
1 0 0 0
0 0 0 1
0 1 0 0
0 0 0 0
· x2 +
1 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
[b] =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
all of the following rules can be deleted.
a__f(X1,X2) → f(X1,X2) (6)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a__f#(X,X) → a__f#(a,b) (8)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.