(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
T(N) → Q(N)
Q(s(X)) → S(p(q(X), d(X)))
Q(s(X)) → P(q(X), d(X))
Q(s(X)) → Q(X)
Q(s(X)) → D(X)
D(s(X)) → S(s(d(X)))
D(s(X)) → S(d(X))
D(s(X)) → D(X)
P(s(X), s(Y)) → S(s(p(X, Y)))
P(s(X), s(Y)) → S(p(X, Y))
P(s(X), s(Y)) → P(X, Y)
F(s(X), cs(Y, Z)) → A(Z)
A(nt(X)) → T(a(X))
A(nt(X)) → A(X)
A(ns(X)) → S(a(X))
A(ns(X)) → A(X)
A(nf(X1, X2)) → F(a(X1), a(X2))
A(nf(X1, X2)) → A(X1)
A(nf(X1, X2)) → A(X2)
The TRS R consists of the following rules:
t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 10 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(s(X), s(Y)) → P(X, Y)
The TRS R consists of the following rules:
t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(s(X), s(Y)) → P(X, Y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- P(s(X), s(Y)) → P(X, Y)
The graph contains the following edges 1 > 1, 2 > 2
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(s(X)) → D(X)
The TRS R consists of the following rules:
t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(s(X)) → D(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- D(s(X)) → D(X)
The graph contains the following edges 1 > 1
(14) YES
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
Q(s(X)) → Q(X)
The TRS R consists of the following rules:
t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
Q(s(X)) → Q(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- Q(s(X)) → Q(X)
The graph contains the following edges 1 > 1
(19) YES
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(nt(X)) → A(X)
A(ns(X)) → A(X)
A(nf(X1, X2)) → F(a(X1), a(X2))
F(s(X), cs(Y, Z)) → A(Z)
A(nf(X1, X2)) → A(X1)
A(nf(X1, X2)) → A(X2)
The TRS R consists of the following rules:
t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
A(nf(X1, X2)) → F(a(X1), a(X2))
A(nf(X1, X2)) → A(X1)
A(nf(X1, X2)) → A(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A(
x1) =
x1
nt(
x1) =
x1
ns(
x1) =
x1
nf(
x1,
x2) =
nf(
x1,
x2)
F(
x1,
x2) =
x2
a(
x1) =
a(
x1)
cs(
x1,
x2) =
x2
t(
x1) =
x1
s(
x1) =
x1
f(
x1,
x2) =
f(
x1,
x2)
0 =
0
nil =
nil
Knuth-Bendix order [KBO] with precedence:
a1 > f2 > nf2
a1 > 0
a1 > nil
and weight map:
0=2
a_1=0
nf_2=2
f_2=2
nil=4
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
s(X) → ns(X)
t(N) → cs(r(q(N)), nt(ns(N)))
t(X) → nt(X)
f(0, X) → nil
f(X1, X2) → nf(X1, X2)
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(nt(X)) → A(X)
A(ns(X)) → A(X)
F(s(X), cs(Y, Z)) → A(Z)
The TRS R consists of the following rules:
t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(ns(X)) → A(X)
A(nt(X)) → A(X)
The TRS R consists of the following rules:
t(N) → cs(r(q(N)), nt(ns(N)))
q(0) → 0
q(s(X)) → s(p(q(X), d(X)))
d(0) → 0
d(s(X)) → s(s(d(X)))
p(0, X) → X
p(X, 0) → X
p(s(X), s(Y)) → s(s(p(X, Y)))
f(0, X) → nil
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z)))
t(X) → nt(X)
s(X) → ns(X)
f(X1, X2) → nf(X1, X2)
a(nt(X)) → t(a(X))
a(ns(X)) → s(a(X))
a(nf(X1, X2)) → f(a(X1), a(X2))
a(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(ns(X)) → A(X)
A(nt(X)) → A(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- A(ns(X)) → A(X)
The graph contains the following edges 1 > 1
- A(nt(X)) → A(X)
The graph contains the following edges 1 > 1
(28) YES