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, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1
Q is empty.
↳ QTRS
↳ DirectTerminationProof
Q restricted rewrite system:
The TRS R consists of the following rules:
*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1
Q is empty.
We use [23] with the following order to prove termination.
Recursive path order with status [2].
Precedence:
i1 > *2 > k2 > 1
Status:
k2: multiset
i1: [1]
1: multiset
*2: [2,1]