(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(quot(x, s(z), s(z)))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → QUOT(x, s(z), s(z))
The TRS R consists of the following rules:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(quot(x, s(z), s(z)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
QUOT(s(x), s(y), z) → QUOT(x, y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(QUOT(x1, x2, x3)) = x1
POL(s(x1)) = 1 + x1
The following usable rules [FROCOS05] were oriented:
none
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOT(x, 0, s(z)) → QUOT(x, s(z), s(z))
The TRS R consists of the following rules:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(quot(x, s(z), s(z)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(6) TRUE