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

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

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

The TRS R 2 is

f(true, x, y, z) → f(and(gt(x, y), gt(x, z)), x, s(y), z)
f(true, x, y, z) → f(and(gt(x, y), gt(x, z)), x, y, s(z))

The signature Sigma is {f}

(2) Obligation:

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

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

(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, z) → F(and(gt(x, y), gt(x, z)), x, s(y), z)
F(true, x, y, z) → AND(gt(x, y), gt(x, z))
F(true, x, y, z) → GT(x, y)
F(true, x, y, z) → GT(x, z)
F(true, x, y, z) → F(and(gt(x, y), gt(x, z)), x, y, s(z))
GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 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, z) → f(and(gt(x, y), gt(x, z)), x, s(y), z)
f(true, x, y, z) → f(and(gt(x, y), gt(x, z)), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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)  =  x1
s(x1)  =  s(x1)
f(x1, x2, x3, x4)  =  x2
true  =  true
and(x1, x2)  =  and(x1, x2)
gt(x1, x2)  =  gt(x1)
0  =  0
false  =  false

Recursive Path Order [RPO].
Precedence:

[s₁, true] > [gt₁, 0, false]
and₂ > [gt₁, 0, false]

The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

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

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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:

F(true, x, y, z) → F(and(gt(x, y), gt(x, z)), x, y, s(z))
F(true, x, y, z) → F(and(gt(x, y), gt(x, z)), x, s(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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