Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
Q is empty.
↳ QTRS
↳ DirectTerminationProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
Q is empty.
We use [23] with the following order to prove termination.
Recursive path order with status [2].
Precedence:
nil > 0 > s1
cons2 > *2 > +2 > s1
cons2 > *2 > 0 > s1
cons2 > sum1 > s1
cons2 > prod1 > 0 > s1
Status:
sum1: multiset
+2: [1,2]
s1: [1]
0: multiset
prod1: multiset
nil: multiset
cons2: multiset
*2: [1,2]