Termination w.r.t. Q proof of /tmp/tmp

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

0 QTRS

↳1 AAECC Innermost (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

(0) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
min(0, y) → 0

The TRS R 2 is

f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The signature Sigma is {f}

(2) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(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:

MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
TWICE(s(x)) → TWICE(x)
-¹(s(x), s(y)) → -¹(x, y)
F(s(x), s(y)) → F(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))
F(s(x), s(y)) → -¹(max(s(x), s(y)), min(s(x), s(y)))
F(s(x), s(y)) → MAX(s(x), s(y))
F(s(x), s(y)) → MIN(s(x), s(y))
F(s(x), s(y)) → P(twice(min(x, y)))
F(s(x), s(y)) → TWICE(min(x, y))
F(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(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 5 SCCs with 6 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:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(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) = x1
s(x1) = s(x1)
min(x1, x2) = x2
0 = 0
max(x1, x2) = max(x1, x2)
twice(x1) = twice(x1)
-(x1, x2) = x1
p(x1) = p(x1)
f(x1, x2) = f

Recursive Path Order [RPO].
Precedence:

max₂ > s₁ > 0
twice₁ > s₁ > 0
f > s₁ > 0
f > p₁ > 0

The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

(9) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(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:

TWICE(s(x)) → TWICE(x)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(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.

TWICE(s(x)) → TWICE(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TWICE(x1) = TWICE(x1)
s(x1) = s(x1)
min(x1, x2) = x2
0 = 0
max(x1, x2) = max(x1, x2)
twice(x1) = twice(x1)
-(x1, x2) = x1
p(x1) = x1
f(x1, x2) = f

Recursive Path Order [RPO].
Precedence:

TWICE₁ > 0
max₂ > s₁ > 0
f > twice₁ > s₁ > 0

The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

(14) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(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:

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(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.

MAX(s(x), s(y)) → MAX(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MAX(x1, x2) = x1
s(x1) = s(x1)
min(x1, x2) = x2
0 = 0
max(x1, x2) = max(x1, x2)
twice(x1) = twice(x1)
-(x1, x2) = x1
p(x1) = p(x1)
f(x1, x2) = f

Recursive Path Order [RPO].
Precedence:

max₂ > s₁ > 0
twice₁ > s₁ > 0
f > s₁ > 0
f > p₁ > 0

The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

(19) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(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

(22) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MIN(s(x), s(y)) → MIN(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2) = x1
s(x1) = s(x1)
min(x1, x2) = x2
0 = 0
max(x1, x2) = max(x1, x2)
twice(x1) = twice(x1)
-(x1, x2) = x1
p(x1) = p(x1)
f(x1, x2) = f

Recursive Path Order [RPO].
Precedence:

max₂ > s₁ > 0
twice₁ > s₁ > 0
f > s₁ > 0
f > p₁ > 0

The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

(24) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

F(s(x), s(y)) → F(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
twice(0) → 0
twice(s(x)) → s(s(twice(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), s(y)) → f(-(max(s(x), s(y)), min(s(x), s(y))), p(twice(min(x, y))))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
twice(0)
twice(s(x0))
-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), s(x1))

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