Termination w.r.t. Q proof of /tmp/tmpkvajIh/cade10.trs

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

(0) Obligation:

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

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The TRS R 2 is

f(true, x, y) → f(gt(x, y), trunc(x), s(y))

The signature Sigma is {f}

(2) Obligation:

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

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
gt(0, x0)
gt(s(x0), 0)
gt(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:

F(true, x, y) → F(gt(x, y), trunc(x), s(y))
F(true, x, y) → GT(x, y)
F(true, x, y) → TRUNC(x)
TRUNC(s(s(x))) → TRUNC(x)
GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
gt(0, x0)
gt(s(x0), 0)
gt(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 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
gt(0, x0)
gt(s(x0), 0)
gt(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.

GT(s(u), s(v)) → GT(u, v)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GT(x1, x2) = GT(x2)
s(x1) = s(x1)
f(x1, x2, x3) = f(x1)
true = true
gt(x1, x2) = gt
trunc(x1) = trunc(x1)
0 = 0
false = false

Recursive path order with status [RPO].
Quasi-Precedence:

trunc₁ > [s₁, f₁, true, gt] > 0
trunc₁ > [s₁, f₁, true, gt] > false

Status:

GT₁: multiset
s₁: multiset
f₁: multiset
true: multiset
gt: []
trunc₁: [1]
0: multiset
false: multiset

The following usable rules [FROCOS05] were oriented:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

(9) Obligation:

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

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
gt(0, x0)
gt(s(x0), 0)
gt(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:

TRUNC(s(s(x))) → TRUNC(x)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
gt(0, x0)
gt(s(x0), 0)
gt(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.

TRUNC(s(s(x))) → TRUNC(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TRUNC(x1) = TRUNC(x1)
s(x1) = s(x1)
f(x1, x2, x3) = f(x2)
true = true
gt(x1, x2) = gt
trunc(x1) = x1
0 = 0
false = false

Recursive path order with status [RPO].
Quasi-Precedence:

TRUNC₁ > [s₁, f₁, false]
gt > true > [s₁, f₁, false]
0 > true > [s₁, f₁, false]

Status:

TRUNC₁: multiset
s₁: [1]
f₁: [1]
true: multiset
gt: multiset
0: multiset
false: multiset

The following usable rules [FROCOS05] were oriented:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

(14) Obligation:

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

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
gt(0, x0)
gt(s(x0), 0)
gt(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:

F(true, x, y) → F(gt(x, y), trunc(x), s(y))

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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