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:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

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:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

(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:

1024¹ → 1024_1¹(0)
1024_1¹(x) → IF(lt(x, 10), x)
1024_1¹(x) → LT(x, 10)
1024_1¹(x) → 10¹
IF(true, x) → DOUBLE(1024_1(s(x)))
IF(true, x) → 1024_1¹(s(x))
LT(s(x), s(y)) → LT(x, y)
DOUBLE(s(x)) → DOUBLE(x)
10¹ → DOUBLE(s(double(s(s(0)))))
10¹ → DOUBLE(s(s(0)))

The TRS R consists of the following rules:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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.

DOUBLE(s(x)) → DOUBLE(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:

s₁ > DOUBLE₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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.

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)  =  LT(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > LT₁

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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(true, x) → 1024_1¹(s(x))
1024_1¹(x) → IF(lt(x, 10), x)

The TRS R consists of the following rules:

1024 → 1024_1(0)
1024_1(x) → if(lt(x, 10), x)
if(true, x) → double(1024_1(s(x)))
if(false, x) → s(0)
lt(0, s(y)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
10 → double(s(double(s(s(0)))))

The set Q consists of the following terms:

1024
1024_1(x0)
if(true, x0)
if(false, x0)
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
double(0)
double(s(x0))
10

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