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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
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:

-¹(s(x), s(y)) → -¹(x, y)
<=¹(s(x), s(y)) → <=¹(x, y)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
F(s(x), 0, z, u) → -¹(z, s(x))
F(s(x), s(y), z, u) → IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) → <=¹(x, y)
F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) → -¹(y, x)
F(s(x), s(y), z, u) → F(x, u, z, u)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
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 3 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

<=¹(s(x), s(y)) → <=¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

<=¹(s(x), s(y)) → <=¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
<=¹(x1, x2)  =  x2
s(x1)  =  s(x1)
-(x1, x2)  =  x1
0  =  0
<=(x1, x2)  =  <=(x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x2, x3)
perfectp(x1)  =  perfectp(x1)
f(x1, x2, x3, x4)  =  f(x1, x2, x4)

Lexicographic Path Order [LPO].
Precedence:

[<=₁, perfectp₁, f₃] > s₁ > false > 0
[<=₁, perfectp₁, f₃] > s₁ > if₂ > 0
[<=₁, perfectp₁, f₃] > true > 0

The following usable rules [FROCOS05] were oriented:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

(9) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
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.

(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:

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

-¹(s(x), s(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2)  =  -¹(x1)
s(x1)  =  s(x1)
-(x1, x2)  =  x1
0  =  0
<=(x1, x2)  =  <=(x1)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x2, x3)
perfectp(x1)  =  perfectp(x1)
f(x1, x2, x3, x4)  =  f(x1, x2, x4)

Lexicographic Path Order [LPO].
Precedence:

[<=₁, true, false, perfectp₁, f₃] > s₁ > 0
[<=₁, true, false, perfectp₁, f₃] > s₁ > if₂

The following usable rules [FROCOS05] were oriented:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

(14) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
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.

(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:

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

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
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.

(18) 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(s(x), -(y, x), z, u)
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, 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, x2, x4)
s(x1)  =  s(x1)
-(x1, x2)  =  -(x1)
0  =  0
<=(x1, x2)  =  <=(x2)
true  =  true
false  =  false
if(x1, x2, x3)  =  if(x2, x3)
perfectp(x1)  =  perfectp(x1)
f(x1, x2, x3, x4)  =  f(x1, x2, x4)

Lexicographic Path Order [LPO].
Precedence:

<=₁ > [true, false, if₂]
perfectp₁ > [F₃, s₁] > -₁ > [true, false, if₂]
perfectp₁ > [F₃, s₁] > [0, f₃] > [true, false, if₂]

The following usable rules [FROCOS05] were oriented:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

(19) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
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, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE