Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

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:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

(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:

MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
DIV(x, y) → QUOT(x, y, 0)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))
QUOT(x, y, z) → ZERO(x)
QUOT(x, y, z) → PLUS(z, s(0))
IF(true, x, y, z) → P(z)
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
IF(false, x, s(y), z) → MINUS(x, s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PLUS(s(x), y) → PLUS(x, s(y))

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

Recursive Path Order [RPO].
Precedence:

[PLUS₁, s₁]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

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

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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