Termination w.r.t. Q proof of /tmp/tmpO8sTcN/19.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 AAECC Innermost (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

(0) Obligation:

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

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, p(y), z)
cond3(false, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
or(false, false) → false
or(true, x) → true
or(x, true) → true
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)
or(false, false) → false
or(true, x) → true
or(x, true) → true
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, p(y), z)
cond3(false, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, y, z)

The signature Sigma is {cond1, cond3, cond2}

(2) Obligation:

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

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, p(y), z)
cond3(false, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
or(false, false) → false
or(true, x) → true
or(x, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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, z) → COND2(gr(x, 0), x, y, z)
COND1(true, x, y, z) → GR(x, 0)
COND2(true, x, y, z) → COND1(or(gr(x, z), gr(y, z)), p(x), y, z)
COND2(true, x, y, z) → OR(gr(x, z), gr(y, z))
COND2(true, x, y, z) → GR(x, z)
COND2(true, x, y, z) → GR(y, z)
COND2(true, x, y, z) → P(x)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND2(false, x, y, z) → GR(y, 0)
COND3(true, x, y, z) → COND1(or(gr(x, z), gr(y, z)), x, p(y), z)
COND3(true, x, y, z) → OR(gr(x, z), gr(y, z))
COND3(true, x, y, z) → GR(x, z)
COND3(true, x, y, z) → GR(y, z)
COND3(true, x, y, z) → P(y)
COND3(false, x, y, z) → COND1(or(gr(x, z), gr(y, z)), x, y, z)
COND3(false, x, y, z) → OR(gr(x, z), gr(y, z))
COND3(false, x, y, z) → GR(x, z)
COND3(false, x, y, z) → GR(y, z)
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, p(y), z)
cond3(false, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
or(false, false) → false
or(true, x) → true
or(x, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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 13 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, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, p(y), z)
cond3(false, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
or(false, false) → false
or(true, x) → true
or(x, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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) = GR(x1)
s(x1) = s(x1)

Lexicographic path order with status [LPO].
Precedence:

s₁ > GR₁

Status:

GR₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, p(y), z)
cond3(false, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
or(false, false) → false
or(true, x) → true
or(x, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
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, z) → COND1(or(gr(x, z), gr(y, z)), p(x), y, z)
COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(true, x, y, z) → COND1(or(gr(x, z), gr(y, z)), x, p(y), z)
COND3(false, x, y, z) → COND1(or(gr(x, z), gr(y, z)), x, y, z)

The TRS R consists of the following rules:

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, p(y), z)
cond3(false, x, y, z) → cond1(or(gr(x, z), gr(y, z)), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
or(false, false) → false
or(true, x) → true
or(x, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
or(false, false)
or(true, x0)
or(x0, true)
p(0)
p(s(x0))

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