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:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

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:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(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:

LEQ(s(x), s(y)) → LEQ(x, y)
-¹(s(x), s(y)) → -¹(x, y)
MOD(s(x), s(y)) → IF(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))
MOD(s(x), s(y)) → LEQ(y, x)
MOD(s(x), s(y)) → MOD(-(s(x), s(y)), s(y))
MOD(s(x), s(y)) → -¹(s(x), s(y))

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(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:

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(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.

-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)

Lexicographic path order with status [LPO].
Precedence:

s₁ > -^1₁

Status:

-^1₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(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:

LEQ(s(x), s(y)) → LEQ(x, y)

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(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.

LEQ(s(x), s(y)) → LEQ(x, y)

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

Lexicographic path order with status [LPO].
Precedence:

s₁ > LEQ₁

Status:

LEQ₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(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:

MOD(s(x), s(y)) → MOD(-(s(x), s(y)), s(y))

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))

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