Certification Problem

Input (TPDB TRS_Standard/SK90/2.41)

The rewrite relation of the following TRS is considered.

norm(nil) 0 (1)
norm(g(x,y)) s(norm(x)) (2)
f(x,nil) g(nil,x) (3)
f(x,g(y,z)) g(f(x,y),z) (4)
rem(nil,y) nil (5)
rem(g(x,y),0) g(x,y) (6)
rem(g(x,y),s(z)) rem(x,z) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(nil) = 0 weight(nil) = 2
prec(0) = 5 weight(0) = 4
prec(norm) = 6 weight(norm) = 2
prec(s) = 2 weight(s) = 1
prec(g) = 1 weight(g) = 0
prec(f) = 4 weight(f) = 0
prec(rem) = 3 weight(rem) = 0
all of the following rules can be deleted.
norm(nil) 0 (1)
norm(g(x,y)) s(norm(x)) (2)
f(x,nil) g(nil,x) (3)
f(x,g(y,z)) g(f(x,y),z) (4)
rem(nil,y) nil (5)
rem(g(x,y),0) g(x,y) (6)
rem(g(x,y),s(z)) rem(x,z) (7)

1.1 R is empty

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