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
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

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
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

(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)
LT(s(x), s(y)) → LT(x, y)
DIV(s(x), s(y)) → IF(lt(x, y), 0, s(div(-(x, y), s(y))))
DIV(s(x), s(y)) → LT(x, y)
DIV(s(x), s(y)) → DIV(-(x, y), s(y))
DIV(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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.

LT(s(x), s(y)) → LT(x, y)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[lt₁, div₂] > s₁ > if₂
[lt₁, div₂] > 0 > if₂
[lt₁, div₂] > false > if₂
[lt₁, div₂] > true > if₂

Status:

if₂: [2,1]
div₂: [1,2]
true: []
false: []
s₁: [1]
lt₁: [1]
0: []

The following usable rules [FROCOS05] were oriented:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

(9) Obligation:

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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)  =  x2
s(x1)  =  s(x1)
-(x1, x2)  =  x1
0  =  0
lt(x1, x2)  =  lt(x1, x2)
false  =  false
true  =  true
if(x1, x2, x3)  =  if(x2, x3)
div(x1, x2)  =  div(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

lt₂ > [s₁, true]
lt₂ > false
div₂ > 0 > [s₁, true]
div₂ > if₂

Status:

if₂: [2,1]
div₂: [1,2]
true: []
false: []
s₁: [1]
lt₂: [2,1]
0: []

The following usable rules [FROCOS05] were oriented:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

(14) Obligation:

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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:

DIV(s(x), s(y)) → DIV(-(x, y), s(y))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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.

DIV(s(x), s(y)) → DIV(-(x, y), s(y))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[0, div₁] > s₁ > if₂ > [DIV₂, -₁]
lt₂ > false > [DIV₂, -₁]
lt₂ > true > [DIV₂, -₁]

Status:

-₁: [1]
if₂: [2,1]
DIV₂: [1,2]
true: []
div₁: [1]
false: []
s₁: [1]
lt₂: [2,1]
0: []

The following usable rules [FROCOS05] were oriented:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

(19) Obligation:

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
if(true, x, y) → x
if(false, x, y) → y
div(x, 0) → 0
div(0, y) → 0
div(s(x), s(y)) → if(lt(x, y), 0, s(div(-(x, y), s(y))))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
div(x0, 0)
div(0, x0)
div(s(x0), s(x1))

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