Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRSToCSRProof (⇔)
↳2 CSR
↳3 CSRInnermostProof (⇔)
↳4 CSR
↳5 CSDependencyPairsProof (⇔)
↳6 QCSDP
↳7 QCSDependencyGraphProof (⇔)
↳8 AND
  ↳9 QCSDP
    ↳10 QCSDPSubtermProof (⇔)
    ↳11 QCSDP
    ↳12 PIsEmptyProof (⇔)
    ↳13 TRUE
  ↳14 QCSDP
    ↳15 QCSDPSubtermProof (⇔)
    ↳16 QCSDP
    ↳17 PIsEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QCSDP
    ↳20 QCSDPSubtermProof (⇔)
    ↳21 QCSDP
    ↳22 PIsEmptyProof (⇔)
    ↳23 TRUE
  ↳24 QCSDP
    ↳25 QCSDPSubtermProof (⇔)
    ↳26 QCSDP
    ↳27 PIsEmptyProof (⇔)
    ↳28 TRUE

(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(2ndspos(0, Z)) → mark(rnil)
active(2ndspos(s(N), cons(X, cons(Y, Z)))) → mark(rcons(posrecip(Y), 2ndsneg(N, Z)))
active(2ndsneg(0, Z)) → mark(rnil)
active(2ndsneg(s(N), cons(X, cons(Y, Z)))) → mark(rcons(negrecip(Y), 2ndspos(N, Z)))
active(pi(X)) → mark(2ndspos(X, from(0)))
active(plus(0, Y)) → mark(Y)
active(plus(s(X), Y)) → mark(s(plus(X, Y)))
active(times(0, Y)) → mark(0)
active(times(s(X), Y)) → mark(plus(Y, times(X, Y)))
active(square(X)) → mark(times(X, X))
active(s(X)) → s(active(X))
active(posrecip(X)) → posrecip(active(X))
active(negrecip(X)) → negrecip(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(rcons(X1, X2)) → rcons(active(X1), X2)
active(rcons(X1, X2)) → rcons(X1, active(X2))
active(from(X)) → from(active(X))
active(2ndspos(X1, X2)) → 2ndspos(active(X1), X2)
active(2ndspos(X1, X2)) → 2ndspos(X1, active(X2))
active(2ndsneg(X1, X2)) → 2ndsneg(active(X1), X2)
active(2ndsneg(X1, X2)) → 2ndsneg(X1, active(X2))
active(pi(X)) → pi(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(times(X1, X2)) → times(active(X1), X2)
active(times(X1, X2)) → times(X1, active(X2))
active(square(X)) → square(active(X))
s(mark(X)) → mark(s(X))
posrecip(mark(X)) → mark(posrecip(X))
negrecip(mark(X)) → mark(negrecip(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
rcons(mark(X1), X2) → mark(rcons(X1, X2))
rcons(X1, mark(X2)) → mark(rcons(X1, X2))
from(mark(X)) → mark(from(X))
2ndspos(mark(X1), X2) → mark(2ndspos(X1, X2))
2ndspos(X1, mark(X2)) → mark(2ndspos(X1, X2))
2ndsneg(mark(X1), X2) → mark(2ndsneg(X1, X2))
2ndsneg(X1, mark(X2)) → mark(2ndsneg(X1, X2))
pi(mark(X)) → mark(pi(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
times(mark(X1), X2) → mark(times(X1, X2))
times(X1, mark(X2)) → mark(times(X1, X2))
square(mark(X)) → mark(square(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(posrecip(X)) → posrecip(proper(X))
proper(negrecip(X)) → negrecip(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(rnil) → ok(rnil)
proper(rcons(X1, X2)) → rcons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(2ndspos(X1, X2)) → 2ndspos(proper(X1), proper(X2))
proper(2ndsneg(X1, X2)) → 2ndsneg(proper(X1), proper(X2))
proper(pi(X)) → pi(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(times(X1, X2)) → times(proper(X1), proper(X2))
proper(square(X)) → square(proper(X))
s(ok(X)) → ok(s(X))
posrecip(ok(X)) → ok(posrecip(X))
negrecip(ok(X)) → ok(negrecip(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
rcons(ok(X1), ok(X2)) → ok(rcons(X1, X2))
from(ok(X)) → ok(from(X))
2ndspos(ok(X1), ok(X2)) → ok(2ndspos(X1, X2))
2ndsneg(ok(X1), ok(X2)) → ok(2ndsneg(X1, X2))
pi(ok(X)) → ok(pi(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
times(ok(X1), ok(X2)) → ok(times(X1, X2))
square(ok(X)) → ok(square(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

(1) QTRSToCSRProof (EQUIVALENT transformation)

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

active(from(X)) → mark(cons(X, from(s(X))))
active(2ndspos(0, Z)) → mark(rnil)
active(2ndspos(s(N), cons(X, cons(Y, Z)))) → mark(rcons(posrecip(Y), 2ndsneg(N, Z)))
active(2ndsneg(0, Z)) → mark(rnil)
active(2ndsneg(s(N), cons(X, cons(Y, Z)))) → mark(rcons(negrecip(Y), 2ndspos(N, Z)))
active(pi(X)) → mark(2ndspos(X, from(0)))
active(plus(0, Y)) → mark(Y)
active(plus(s(X), Y)) → mark(s(plus(X, Y)))
active(times(0, Y)) → mark(0)
active(times(s(X), Y)) → mark(plus(Y, times(X, Y)))
active(square(X)) → mark(times(X, X))
active(s(X)) → s(active(X))
active(posrecip(X)) → posrecip(active(X))
active(negrecip(X)) → negrecip(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(rcons(X1, X2)) → rcons(active(X1), X2)
active(rcons(X1, X2)) → rcons(X1, active(X2))
active(from(X)) → from(active(X))
active(2ndspos(X1, X2)) → 2ndspos(active(X1), X2)
active(2ndspos(X1, X2)) → 2ndspos(X1, active(X2))
active(2ndsneg(X1, X2)) → 2ndsneg(active(X1), X2)
active(2ndsneg(X1, X2)) → 2ndsneg(X1, active(X2))
active(pi(X)) → pi(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(times(X1, X2)) → times(active(X1), X2)
active(times(X1, X2)) → times(X1, active(X2))
active(square(X)) → square(active(X))
s(mark(X)) → mark(s(X))
posrecip(mark(X)) → mark(posrecip(X))
negrecip(mark(X)) → mark(negrecip(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
rcons(mark(X1), X2) → mark(rcons(X1, X2))
rcons(X1, mark(X2)) → mark(rcons(X1, X2))
from(mark(X)) → mark(from(X))
2ndspos(mark(X1), X2) → mark(2ndspos(X1, X2))
2ndspos(X1, mark(X2)) → mark(2ndspos(X1, X2))
2ndsneg(mark(X1), X2) → mark(2ndsneg(X1, X2))
2ndsneg(X1, mark(X2)) → mark(2ndsneg(X1, X2))
pi(mark(X)) → mark(pi(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
times(mark(X1), X2) → mark(times(X1, X2))
times(X1, mark(X2)) → mark(times(X1, X2))
square(mark(X)) → mark(square(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(posrecip(X)) → posrecip(proper(X))
proper(negrecip(X)) → negrecip(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(rnil) → ok(rnil)
proper(rcons(X1, X2)) → rcons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(2ndspos(X1, X2)) → 2ndspos(proper(X1), proper(X2))
proper(2ndsneg(X1, X2)) → 2ndsneg(proper(X1), proper(X2))
proper(pi(X)) → pi(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(times(X1, X2)) → times(proper(X1), proper(X2))
proper(square(X)) → square(proper(X))
s(ok(X)) → ok(s(X))
posrecip(ok(X)) → ok(posrecip(X))
negrecip(ok(X)) → ok(negrecip(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
rcons(ok(X1), ok(X2)) → ok(rcons(X1, X2))
from(ok(X)) → ok(from(X))
2ndspos(ok(X1), ok(X2)) → ok(2ndspos(X1, X2))
2ndsneg(ok(X1), ok(X2)) → ok(2ndsneg(X1, X2))
pi(ok(X)) → ok(pi(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
times(ok(X1), ok(X2)) → ok(times(X1, X2))
square(ok(X)) → ok(square(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
Special symbols used for the transformation (see [GM04]):
top: top, active: active, mark: mark, ok: ok, proper: proper
The replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
2ndspos: {1, 2}
0: empty set
rnil: empty set
rcons: {1, 2}
posrecip: {1}
2ndsneg: {1, 2}
negrecip: {1}
pi: {1}
plus: {1, 2}
times: {1, 2}
square: {1}
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).

(2) Obligation:

Context-sensitive 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, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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 replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
2ndspos: {1, 2}
0: empty set
rnil: empty set
rcons: {1, 2}
posrecip: {1}
2ndsneg: {1, 2}
negrecip: {1}
pi: {1}
plus: {1, 2}
times: {1, 2}
square: {1}

(3) CSRInnermostProof (EQUIVALENT transformation)

The CSR is orthogonal. By [CS_Inn] we can switch to innermost.

(4) Obligation:

Context-sensitive 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, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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 replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
2ndspos: {1, 2}
0: empty set
rnil: empty set
rcons: {1, 2}
posrecip: {1}
2ndsneg: {1, 2}
negrecip: {1}
pi: {1}
plus: {1, 2}
times: {1, 2}
square: {1}

Innermost Strategy.

(5) CSDependencyPairsProof (EQUIVALENT transformation)

Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.

(6) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square, 2NDSNEG, 2NDSPOS, PI, FROM, PLUS, TIMES, SQUARE} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The ordinary context-sensitive dependency pairs DPo are:

2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)
2NDSNEG(s(N), cons(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 collapsing dependency pairs are DPc:

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


The hidden terms of R are:

from(s(x0))

Every hiding context is built from:

s on positions {1}
from on positions {1}

Hence, the new unhiding pairs DPu are :

2NDSPOS(s(N), cons(X, cons(Y, Z))) → U(Y)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → U(Z)
2NDSNEG(s(N), cons(X, cons(Y, Z))) → U(Y)
2NDSNEG(s(N), cons(X, cons(Y, Z))) → U(Z)
U(s(x_0)) → U(x_0)
U(from(x_0)) → U(x_0)
U(from(s(x0))) → FROM(s(x0))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(7) QCSDependencyGraphProof (EQUIVALENT transformation)

The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 4 SCCs with 7 less nodes.
The rules PI(x0) → 2NDSPOS(x0, from(0)) and 2NDSPOS(s(z0), cons(z1, cons(z2, z3))) → 2NDSNEG(z0, z3) form no chain, because ECapµR'(2NDSPOS(s(z0), cons(z1, cons(z2, z3)))) = 2NDSPOS(s(x_1), cons(x_2, cons(z2, z3))) does not unify with 2NDSPOS(x0, from(0)).
R' =

( cons(X, from(s(X))), from(X))


The rules PI(x0) → 2NDSPOS(x0, from(0)) and 2NDSPOS(s(z0), cons(z1, cons(z2, z3))) → U(z2) form no chain, because ECapµR'(2NDSPOS(s(z0), cons(z1, cons(z2, z3)))) = 2NDSPOS(s(x_1), cons(x_2, cons(z2, z3))) does not unify with 2NDSPOS(x0, from(0)).
R' =

( cons(X, from(s(X))), from(X))


The rules PI(x0) → 2NDSPOS(x0, from(0)) and 2NDSPOS(s(z0), cons(z1, cons(z2, z3))) → U(z3) form no chain, because ECapµR'(2NDSPOS(s(z0), cons(z1, cons(z2, z3)))) = 2NDSPOS(s(x_1), cons(x_2, cons(z2, z3))) does not unify with 2NDSPOS(x0, from(0)).
R' =

( cons(X, from(s(X))), from(X))


(8) Complex Obligation (AND)

(9) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The TRS P consists of the following rules:

U(s(x_0)) → U(x_0)
U(from(x_0)) → U(x_0)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(10) QCSDPSubtermProof (EQUIVALENT transformation)

We use the subterm processor [DA_EMMES].


The following pairs can be oriented strictly and are deleted.


U(s(x_0)) → U(x_0)
U(from(x_0)) → U(x_0)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
U(x1)  =  x1

Subterm Order

(11) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(12) PIsEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R,µ)-chain.

(13) TRUE

(14) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square, PLUS} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(15) QCSDPSubtermProof (EQUIVALENT transformation)

We use the subterm processor [DA_EMMES].


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.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  x1

Subterm Order

(16) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(17) PIsEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R,µ)-chain.

(18) TRUE

(19) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square, TIMES} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(20) QCSDPSubtermProof (EQUIVALENT transformation)

We use the subterm processor [DA_EMMES].


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.
none
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  x1

Subterm Order

(21) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(22) PIsEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R,µ)-chain.

(23) TRUE

(24) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square, 2NDSPOS, 2NDSNEG} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:

2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, 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, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(25) QCSDPSubtermProof (EQUIVALENT transformation)

We use the subterm processor [DA_EMMES].


The following pairs can be oriented strictly and are deleted.


2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
2NDSPOS(x1, x2)  =  x1
2NDSNEG(x1, x2)  =  x1

Subterm Order

(26) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, 2ndspos, rcons, posrecip, 2ndsneg, negrecip, pi, plus, times, square} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(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, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(27) PIsEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R,µ)-chain.

(28) TRUE