Termination w.r.t. Q proof of /tmp/tmpXLQ8FT/ExAppendixB

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

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 DependencyGraphProof (⇔)

    ↳21 TRUE

  ↳22 QDP

(0) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(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:

FROM(X) → FROM(s(X))
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, Z)
2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, Z)
PI(X) → 2NDSPOS(X, from(0))
PI(X) → FROM(0)
PLUS(s(X), Y) → PLUS(X, Y)
TIMES(s(X), Y) → PLUS(Y, times(X, Y))
TIMES(s(X), Y) → TIMES(X, Y)
SQUARE(X) → TIMES(X, X)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(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 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(X), Y) → PLUS(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(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.

PLUS(s(X), Y) → PLUS(X, Y)

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)
from(x1) = from(x1)
cons(x1, x2) = cons(x1)
2ndspos(x1, x2) = x2
0 = 0
rnil = rnil
cons2(x1, x2) = cons2(x1)
rcons(x1, x2) = rcons
posrecip(x1) = posrecip
2ndsneg(x1, x2) = 2ndsneg
negrecip(x1) = negrecip
pi(x1) = pi(x1)
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

posrecip > [0, rnil]
negrecip > [0, rnil]
square₁ > times₂ > plus₂ > [PLUS₂, s₁, from₁, cons₁, pi₁] > cons2₁ > [rcons, 2ndsneg] > [0, rnil]

Status:

PLUS₂: [1,2]
s₁: [1]
from₁: [1]
cons₁: [1]
0: []
rnil: []
cons2₁: [1]
rcons: []
posrecip: []
2ndsneg: []
negrecip: []
pi₁: [1]
plus₂: [1,2]
times₂: [2,1]
square₁: [1]

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

(9) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(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:

TIMES(s(X), Y) → TIMES(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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.

TIMES(s(X), Y) → TIMES(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2) = TIMES(x1, x2)
s(x1) = s(x1)
from(x1) = x1
cons(x1, x2) = x1
2ndspos(x1, x2) = 2ndspos(x1, x2)
0 = 0
rnil = rnil
cons2(x1, x2) = x1
rcons(x1, x2) = x1
posrecip(x1) = posrecip
2ndsneg(x1, x2) = 2ndsneg
negrecip(x1) = negrecip
pi(x1) = pi(x1)
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[2ndsneg, negrecip] > [s₁, 2ndspos₂] > TIMES₂ > rnil
[2ndsneg, negrecip] > [s₁, 2ndspos₂] > posrecip > rnil
pi₁ > [s₁, 2ndspos₂] > TIMES₂ > rnil
pi₁ > [s₁, 2ndspos₂] > posrecip > rnil
pi₁ > 0 > rnil
square₁ > times₂ > 0 > rnil
square₁ > times₂ > plus₂ > [s₁, 2ndspos₂] > TIMES₂ > rnil
square₁ > times₂ > plus₂ > [s₁, 2ndspos₂] > posrecip > rnil

Status:

TIMES₂: [2,1]
s₁: [1]
2ndspos₂: [1,2]
0: []
rnil: []
posrecip: []
2ndsneg: []
negrecip: []
pi₁: [1]
plus₂: [2,1]
times₂: [2,1]
square₁: [1]

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

(14) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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:

2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, Z)
2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, Z))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, Z)
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, Z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2NDSPOS(x1, x2) = 2NDSPOS(x1)
s(x1) = s(x1)
cons2(x1, x2) = x2
cons(x1, x2) = x2
2NDSNEG(x1, x2) = x1
from(x1) = from
2ndspos(x1, x2) = 2ndspos(x1, x2)
0 = 0
rnil = rnil
rcons(x1, x2) = rcons(x2)
posrecip(x1) = posrecip
2ndsneg(x1, x2) = 2ndsneg(x1, x2)
negrecip(x1) = negrecip
pi(x1) = pi(x1)
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

posrecip > from
pi₁ > [2ndspos₂, 2ndsneg₂] > rcons₁ > from
pi₁ > 0 > rnil > from
square₁ > times₂ > plus₂ > [s₁, negrecip] > 2NDSPOS₁ > from
square₁ > times₂ > plus₂ > [s₁, negrecip] > [2ndspos₂, 2ndsneg₂] > rcons₁ > from

Status:

2NDSPOS₁: [1]
s₁: [1]
from: []
2ndspos₂: [2,1]
0: []
rnil: []
rcons₁: [1]
posrecip: []
2ndsneg₂: [2,1]
negrecip: []
pi₁: [1]
plus₂: [1,2]
times₂: [2,1]
square₁: [1]

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

(19) Obligation:

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

2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, Z))
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, Z))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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

(20) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(21) TRUE

(22) Obligation:

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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