Certification Problem

Input (TPDB TRS_Standard/SK90/2.23)

The rewrite relation of the following TRS is considered.

fac(0) 1 (1)
fac(s(x)) *(s(x),fac(x)) (2)
floop(0,y) y (3)
floop(s(x),y) floop(x,*(s(x),y)) (4)
*(x,0) 0 (5)
*(x,s(y)) +(*(x,y),x) (6)
+(x,0) x (7)
+(x,s(y)) s(+(x,y)) (8)
1 s(0) (9)
fac(0) s(0) (10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(+) = 1 status(+) = [1, 2] list-extension(+) = Lex
prec(floop) = 3 status(floop) = [1, 2] list-extension(floop) = Lex
prec(*) = 2 status(*) = [2, 1] list-extension(*) = Lex
prec(s) = 0 status(s) = [1] list-extension(s) = Lex
prec(1) = 0 status(1) = [] list-extension(1) = Lex
prec(fac) = 3 status(fac) = [1] list-extension(fac) = Lex
prec(0) = 0 status(0) = [] list-extension(0) = Lex
and the following Max-polynomial interpretation
[+(x1, x2)] = max(0, 0 + 1 · x1, 2 + 1 · x2)
[floop(x1, x2)] = max(2, 2 + 1 · x1, 0 + 1 · x2)
[*(x1, x2)] = max(0, 2 + 1 · x1, 0 + 1 · x2)
[s(x1)] = max(2, 0 + 1 · x1)
[1] = max(7)
[fac(x1)] = max(3, 2 + 1 · x1)
[0] = 6
all of the following rules can be deleted.
fac(0) 1 (1)
fac(s(x)) *(s(x),fac(x)) (2)
floop(0,y) y (3)
floop(s(x),y) floop(x,*(s(x),y)) (4)
*(x,0) 0 (5)
*(x,s(y)) +(*(x,y),x) (6)
+(x,0) x (7)
+(x,s(y)) s(+(x,y)) (8)
1 s(0) (9)
fac(0) s(0) (10)

1.1 R is empty

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