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(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, 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(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, x2)
plus(x0, 0)
plus(x0, s(x1))
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, z) → F(gt(x, plus(y, z)), x, s(y), z)
F(true, x, y, z) → GT(x, plus(y, z))
F(true, x, y, z) → PLUS(y, z)
F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
PLUS(n, s(m)) → PLUS(n, m)
GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, x2)
plus(x0, 0)
plus(x0, s(x1))
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, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, x2)
plus(x0, 0)
plus(x0, s(x1))
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)  =  x1
s(x1)  =  s(x1)
f(x1, x2, x3, x4)  =  f(x1, x2)
true  =  true
gt(x1, x2)  =  gt
plus(x1, x2)  =  plus(x1, x2)
0  =  0
false  =  false

Recursive Path Order [RPO].
Precedence:

plus₂ > s₁ > [f₂, true, gt, 0] > false

The following usable rules [FROCOS05] were oriented:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, x2)
plus(x0, 0)
plus(x0, s(x1))
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:

PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, x2)
plus(x0, 0)
plus(x0, s(x1))
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.

PLUS(n, s(m)) → PLUS(n, m)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x1, x2)
s(x1)  =  s(x1)
f(x1, x2, x3, x4)  =  x1
true  =  true
gt(x1, x2)  =  gt
plus(x1, x2)  =  plus(x1, x2)
0  =  0
false  =  false

Recursive Path Order [RPO].
Precedence:

[true, gt, false] > plus₂ > [PLUS₂, s₁]
0 > [PLUS₂, s₁]

The following usable rules [FROCOS05] were oriented:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, x2)
plus(x0, 0)
plus(x0, s(x1))
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, z) → F(gt(x, plus(y, z)), x, y, s(z))
F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)

The TRS R consists of the following rules:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
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, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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