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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)

The signature Sigma is {cond1, cond2}

(2) Obligation:

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

(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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND1(true, x, y) → GR(y, 0)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(true, x, y) → GR(y, 0)
COND2(true, x, y) → P(y)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)
COND2(false, x, y) → GR(x, 0)
COND2(false, x, y) → P(x)
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

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 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

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.

GR(s(x), s(y)) → GR(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GR(x1, x2)  =  x2
s(x1)  =  s(x1)
cond1(x1, x2, x3)  =  cond1(x2, x3)
true  =  true
cond2(x1, x2, x3)  =  cond2(x2, x3)
gr(x1, x2)  =  gr
0  =  0
p(x1)  =  x1
false  =  false

Recursive Path Order [RPO].
Precedence:

s₁ > 0
[cond1₂, true, cond2₂, gr, false] > 0

The following usable rules [FROCOS05] were oriented:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

(9) Obligation:

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

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:

COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)
COND1(true, x, y) → COND2(gr(y, 0), x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

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