(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The TRS R 2 is
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
(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:
F(true, xs) → F(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
F(true, xs) → EQ(s(length(xs)), length(cons(a, xs)))
F(true, xs) → LENGTH(xs)
F(true, xs) → LENGTH(cons(a, xs))
LENGTH(cons(x, xs)) → LENGTH(xs)
EQ(s(x), s(y)) → EQ(x, y)
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
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 3 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:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(s(x), s(y)) → EQ(x, y)
R is empty.
The set Q consists of the following terms:
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(s(x), s(y)) → EQ(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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:
- EQ(s(x), s(y)) → EQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(x, xs)) → LENGTH(xs)
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(x, xs)) → LENGTH(xs)
R is empty.
The set Q consists of the following terms:
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(17) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(x, xs)) → LENGTH(xs)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) 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:
- LENGTH(cons(x, xs)) → LENGTH(xs)
The graph contains the following edges 1 > 1
(20) YES
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, xs) → F(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
The set Q consists of the following terms:
f(true, x0)
length(nil)
length(cons(x0, x1))
eq(0, 0)
eq(s(x0), 0)
eq(0, s(x0))
eq(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(22) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, xs) → F(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
Q is empty.
We have to consider all (P,Q,R)-chains.
(24) NonLoopProof (COMPLETE transformation)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 1,
σ' = [ ], and μ' = [ ] on the rule
F(
true,
cons(
a,
zr1))[
zr1 /
cons(
a,
zr1)]
n[
zr1 /
nil] →
F(
true,
cons(
a,
cons(
a,
zr1)))[
zr1 /
cons(
a,
zr1)]
n[
zr1 /
nil]
This rule is correct for the QDP as the following derivation shows:
F(
true,
cons(
a,
zr1))[
zr1 /
cons(
a,
zr1)]
n[
zr1 /
nil] →
F(
true,
cons(
a,
cons(
a,
zr1)))[
zr1 /
cons(
a,
zr1)]
n[
zr1 /
nil]
by Equivalence by Domain Renaming of the lhs with [
zl0 /
zr1]
intermediate steps: Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
F(
true,
cons(
x1,
zl1))[
zl1 /
cons(
x1,
zl1)]
n[
zl1 /
nil] →
F(
true,
cons(
a,
cons(
x1,
zr1)))[
zr1 /
cons(
x1,
zr1)]
n[
zr1 /
nil]
by Rewrite t with the rewrite sequence
: [([0,1],length(nil) -> 0), ([0],eq(0, 0) -> true)]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs)
F(true, cons(x1, zl1))[zr1 / cons(x1, zr1), zl1 / cons(x1, zl1)]n[zr1 / nil, zl1 / nil] → F(eq(0, length(nil)), cons(a, cons(x1, zr1)))[zr1 / cons(x1, zr1), zl1 / cons(x1, zl1)]n[zr1 / nil, zl1 / nil]
by Narrowing at position: [0,0]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Equiv Sµ (rhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
F(true, cons(x1, zl1))[zr2 / s(zr2), zr3 / cons(x1, zr3), zr5 / s(zr5), zl1 / cons(x1, zl1)]n[zr2 / length(x0), zr3 / x0, zr5 / length(x0), y1 / length(x0), y0 / length(x0), zl1 / x0] → F(eq(y1, y0), cons(a, cons(x1, zr3)))[zr2 / s(zr2), zr3 / cons(x1, zr3), zr5 / s(zr5), zl1 / cons(x1, zl1)]n[zr2 / length(x0), zr3 / x0, zr5 / length(x0), y1 / length(x0), y0 / length(x0), zl1 / x0]
by Narrowing at position: [0]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
F(true, cons(x1, zl1))[zr2 / s(zr2), zr3 / cons(x1, zr3), zt1 / s(zt1), zl1 / cons(x1, zl1)]n[zr2 / length(y1), zr3 / y1, zt1 / length(y1), zl1 / y1, x0 / y1] → F(eq(s(s(zr2)), s(s(zt1))), cons(a, cons(x1, zr3)))[zr2 / s(zr2), zr3 / cons(x1, zr3), zt1 / s(zt1), zl1 / cons(x1, zl1)]n[zr2 / length(y1), zr3 / y1, zt1 / length(y1), zl1 / y1, x0 / y1]
by Narrowing at position: [0,1,0]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
F(true, cons(x1, zl1))[zr2 / s(zr2), zr3 / cons(x1, zr3), zl1 / cons(x1, zl1)]n[zr2 / length(x0), zr3 / x0, zl1 / x0] → F(eq(s(s(zr2)), s(length(cons(x1, zr3)))), cons(a, cons(x1, zr3)))[zr2 / s(zr2), zr3 / cons(x1, zr3), zl1 / cons(x1, zl1)]n[zr2 / length(x0), zr3 / x0, zl1 / x0]
by Narrowing at position: [0,1]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (lhs)
F(true, cons(y0, zs1))[zt1 / s(zt1), zs1 / cons(y0, zs1)]n[zt1 / length(y1), zs1 / y1] → F(eq(s(s(zt1)), length(cons(a, cons(y0, zs1)))), cons(a, cons(y0, zs1)))[zt1 / s(zt1), zs1 / cons(y0, zs1)]n[zt1 / length(y1), zs1 / y1]
by Narrowing at position: [0,0,0]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate Sigma - Instantiation - Instantiation
F(true, xs)[ ]n[ ] → F(eq(s(length(xs)), length(cons(a, xs))), cons(a, xs))[ ]n[ ]
by Rule from TRS P
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs)
length(cons(x, xs))[xs / cons(x, xs)]n[ ] → s(z)[xs / cons(x, xs), z / s(z)]n[z / length(xs)]
by PatternCreation II with pi: [0], sigma: [xs / cons(x, xs)]
length(cons(x, xs))[ ]n[ ] → s(length(xs))[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate Sigma - Instantiation - Instantiation - Instantiation - Instantiation
length(cons(x, xs))[ ]n[ ] → s(length(xs))[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (lhs) - Instantiation - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs)
length(cons(x, xs))[xs / cons(x, xs)]n[ ] → s(z)[xs / cons(x, xs), z / s(z)]n[z / length(xs)]
by PatternCreation II with pi: [0], sigma: [xs / cons(x, xs)]
length(cons(x, xs))[ ]n[ ] → s(length(xs))[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (lhs) - Equiv DR (lhs) - Expand Sigma - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
eq(s(x), s(y))[x / s(x), y / s(y)]n[ ] → eq(x, y)[ ]n[ ]
by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x), y / s(y)]
eq(s(x), s(y))[ ]n[ ] → eq(x, y)[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
length(nil)[ ]n[ ] → 0[ ]n[ ]
by Rule from TRS R
(25) NO