Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex6_Luc98_Z)

The rewrite relation of the following TRS is considered.

first(0,X) nil (1)
first(s(X),cons(Y,Z)) cons(Y,n__first(X,activate(Z))) (2)
from(X) cons(X,n__from(s(X))) (3)
first(X1,X2) n__first(X1,X2) (4)
from(X) n__from(X) (5)
activate(n__first(X1,X2)) first(X1,X2) (6)
activate(n__from(X)) from(X) (7)
activate(X) X (8)

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
[first(x1, x2)] = 2 · x1 + 2 · x2
[0] = 0
[nil] = 0
[s(x1)] = 1 · x1
[cons(x1, x2)] = 1 + 1 · x1 + 1 · x2
[n__first(x1, x2)] = 1 · x1 + 1 · x2
[activate(x1)] = 2 · x1
[from(x1)] = 2 + 2 · x1
[n__from(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
first(s(X),cons(Y,Z)) cons(Y,n__first(X,activate(Z))) (2)
from(X) n__from(X) (5)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[first(x1, x2)] = 2 + 2 · x1 + 2 · x2
[0] = 0
[nil] = 1
[from(x1)] = 1 + 2 · x1
[cons(x1, x2)] = 1 · x1 + 1 · x2
[n__from(x1)] = 1 · x1
[s(x1)] = 1 · x1
[n__first(x1, x2)] = 2 + 2 · x1 + 1 · x2
[activate(x1)] = 1 + 2 · x1
all of the following rules can be deleted.
first(0,X) nil (1)
from(X) cons(X,n__from(s(X))) (3)
activate(n__first(X1,X2)) first(X1,X2) (6)
activate(X) X (8)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(activate) = 4 weight(activate) = 1
prec(n__from) = 2 weight(n__from) = 1
prec(from) = 3 weight(from) = 2
prec(first) = 1 weight(first) = 0
prec(n__first) = 0 weight(n__first) = 0
all of the following rules can be deleted.
first(X1,X2) n__first(X1,X2) (4)
activate(n__from(X)) from(X) (7)

1.1.1.1 R is empty

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