Certification Problem

Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex9_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(a,X,X) → a__f(X,a__b,b) (1)
a__b → a (2)
mark(f(X1,X2,X3)) → a__f(X1,mark(X2),X3) (3)
mark(b) → a__b (4)
mark(a) → a (5)
a__f(X1,X2,X3) → f(X1,X2,X3) (6)
a__b → b (7)
The evaluation strategy is innermost.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a__f(x1, x2, x3)] = 2 + 2 · x1 + 2 · x2 + 2 · x3
[a] = 1
[a__b] = 1
[b] = 0
[mark(x1)] = 1 + 2 · x1
[f(x1, x2, x3)] = 2 + 2 · x1 + 2 · x2 + 2 · x3
all of the following rules can be deleted.
mark(f(X1,X2,X3)) → a__f(X1,mark(X2),X3) (3)
mark(a) → a (5)
a__b → b (7)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a__f(x1, x2, x3)] = 1 + 1 · x1 + 2 · x2 + 2 · x3
[a] = 0
[a__b] = 0
[b] = 0
[mark(x1)] = 2 + 2 · x1
[f(x1, x2, x3)] = 1 + 1 · x1 + 2 · x2 + 1 · x3
all of the following rules can be deleted.
mark(b) → a__b (4)

1.1.1 Rule Removal

Using the
prec(a__f) = 1 stat(a__f) = mul
prec(a) = 2 stat(a) = mul
prec(a__b) = 2 stat(a__b) = mul
prec(b) = 0 stat(b) = mul
prec(f) = 0 stat(f) = lex

π(a__f) = [1,2,3]
π(a) = []
π(a__b) = []
π(b) = []
π(f) = [2,3,1]

all of the following rules can be deleted.
a__f(X1,X2,X3) → f(X1,X2,X3) (6)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a__f(x1, x2, x3)] = 2 · x1 + 1 · x2 + 2 · x3
[a] = 1
[a__b] = 1
[b] = 0
all of the following rules can be deleted.
a__f(a,X,X) → a__f(X,a__b,b) (1)

1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1 R is empty

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