Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
F(s(x), s(y), z, u) → F(x, u, z, u)

The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)

The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4)  =  F(x1)
s(x1)  =  s(x1)
0  =  0
minus(x1, x2)  =  minus

Recursive Path Order [RPO].
Precedence:

F₁ > s₁ > minus
0 > minus

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE