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

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

The TRS R 2 is

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

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))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
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:

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), p(x), p(y))
COND2(true, x, y) → GR(y, 0)
COND2(true, x, y) → P(x)
COND2(true, x, y) → P(y)
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), x, y)
COND2(false, x, y) → AND(eq(x, y), gr(x, 0))
COND2(false, x, y) → EQ(x, y)
COND2(false, x, y) → GR(x, 0)
GR(s(x), s(y)) → GR(x, y)
EQ(s(x), s(y)) → EQ(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), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), 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
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

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))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
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 3 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

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))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
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.

EQ(s(x), s(y)) → EQ(x, y)

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

Lexicographic path order with status [LPO].
Precedence:

s₁ > EQ₁
s₁ > eq₂ > true
0 > false > and₂
0 > false > eq₂ > true

Status:

eq₂: [1,2]
true: []
false: []
EQ₁: [1]
s₁: [1]
and₂: [1,2]
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), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), 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
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

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

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))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
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:

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

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))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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.

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)  =  GR(x1)
s(x1)  =  s(x1)
cond1(x1, x2, x3)  =  x2
true  =  true
cond2(x1, x2, x3)  =  x2
gr(x1, x2)  =  gr(x2)
0  =  0
p(x1)  =  x1
false  =  false
and(x1, x2)  =  and(x1)
eq(x1, x2)  =  x1

Lexicographic path order with status [LPO].
Precedence:

s₁ > GR₁ > false
s₁ > gr₁ > false
0 > true > gr₁ > false
and₁ > true > gr₁ > false

Status:

GR₁: [1]
gr₁: [1]
true: []
false: []
and₁: [1]
s₁: [1]
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), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), 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
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

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

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))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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:

COND2(true, x, y) → COND2(gr(y, 0), p(x), p(y))
COND2(false, x, y) → COND1(and(eq(x, y), gr(x, 0)), 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), p(x), p(y))
cond2(false, x, y) → cond1(and(eq(x, y), gr(x, 0)), 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
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

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))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)

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