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:

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

[GT₂, s₁]

Status:

s₁: [1]
GT₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(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)  =  x1
s(x1)  =  s(x1)

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

trivial

Status:

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:

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.