(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))), append(cons(a, nil), 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)
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
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:
F(true, xs) → F(eq(s(length(xs)), length(cons(a, xs))), append(cons(a, nil), xs))
F(true, xs) → EQ(s(length(xs)), length(cons(a, xs)))
F(true, xs) → LENGTH(xs)
F(true, xs) → LENGTH(cons(a, xs))
F(true, xs) → APPEND(cons(a, nil), xs)
LENGTH(cons(x, xs)) → LENGTH(xs)
EQ(s(x), s(y)) → EQ(x, y)
APPEND(xs, ys) → APPENDAKK(reverse(xs), ys)
APPEND(xs, ys) → REVERSE(xs)
APPENDAKK(cons(x, xs), ys) → APPENDAKK(xs, cons(x, ys))
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(xs)
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), append(cons(a, nil), 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)
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
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 5 SCCs with 5 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APPENDAKK(cons(x, xs), ys) → APPENDAKK(xs, cons(x, ys))
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), append(cons(a, nil), 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)
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) 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:
- APPENDAKK(cons(x, xs), ys) → APPENDAKK(xs, cons(x, ys))
The graph contains the following edges 1 > 1
(7) YES
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(xs)
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), append(cons(a, nil), 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)
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(xs)
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), append(cons(a, nil), 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)
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
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))
append(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(xs)
The TRS R consists of the following rules:
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
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))
append(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(13) 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))
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(xs)
The TRS R consists of the following rules:
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
The set Q consists of the following terms:
append(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(15) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
REVERSE(cons(x, xs)) → REVERSE(xs)
Used ordering: Polynomial interpretation [POLO]:
POL(APPEND(x1, x2)) = x1 + x2
POL(REVERSE(x1)) = x1
POL(append(x1, x2)) = x1 + x2
POL(appendAkk(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = 1 + x1 + x2
POL(nil) = 0
POL(reverse(x1)) = x1
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
The TRS R consists of the following rules:
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
The set Q consists of the following terms:
append(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(17) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
Used ordering: Polynomial interpretation [POLO]:
POL(APPEND(x1, x2)) = 1 + 2·x1 + x2
POL(REVERSE(x1)) = 2·x1
POL(append(x1, x2)) = x1 + x2
POL(appendAkk(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = 2 + x1 + x2
POL(nil) = 0
POL(reverse(x1)) = x1
(18) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
The set Q consists of the following terms:
append(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(19) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(20) YES
(21) 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))), append(cons(a, nil), 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)
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(22) 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
(23) YES
(24) 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))), append(cons(a, nil), 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)
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) 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
(26) YES
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, xs) → F(eq(s(length(xs)), length(cons(a, xs))), append(cons(a, nil), xs))
The TRS R consists of the following rules:
f(true, xs) → f(eq(s(length(xs)), length(cons(a, xs))), append(cons(a, nil), 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)
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(28) NonLoopProof (EQUIVALENT 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:
intermediate steps: Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
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
intermediate steps: Equivalent (Remove Unused)
F(true, cons(x1, zl1))[zl1 / cons(x1, zl1), zr1 / cons(x1, zr1)]n[zl1 / nil, zr1 / nil] → F(eq(0, length(nil)), append(cons(a, nil), cons(x1, zr1)))[zl1 / cons(x1, zl1), zr1 / cons(x1, zr1)]n[zl1 / nil, zr1 / nil]
by Narrowing at position: [0,0]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Simplify mu) - Equivalent (Remove Unused)
F(true, cons(x1, zl1))[zl1 / cons(x1, zl1), zr2 / cons(x1, zr2), zr3 / s(zr3), zr5 / s(zr5)]n[zl1 / x0, zr2 / x0, zr3 / length(x0), zr5 / length(x0), y1 / length(x0), y0 / length(x0)] → F(eq(y1, y0), append(cons(a, nil), cons(x1, zr2)))[zl1 / cons(x1, zl1), zr2 / cons(x1, zr2), zr3 / s(zr3), zr5 / s(zr5)]n[zl1 / x0, zr2 / x0, zr3 / length(x0), zr5 / length(x0), y1 / length(x0), y0 / length(x0)]
by Narrowing at position: [0]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
F(true, cons(x1, zl1))[zl1 / cons(x1, zl1), zr2 / cons(x1, zr2), zr3 / s(zr3), zt1 / s(zt1)]n[zl1 / y0, x0 / y0, zr2 / y0, zr3 / length(y0), zt1 / length(y0)] → F(eq(s(s(zr3)), s(s(zt1))), append(cons(a, nil), cons(x1, zr2)))[zl1 / cons(x1, zl1), zr2 / cons(x1, zr2), zr3 / s(zr3), zt1 / s(zt1)]n[zl1 / y0, x0 / y0, zr2 / y0, zr3 / length(y0), zt1 / length(y0)]
by Narrowing at position: [0,1,0]
intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
F(true, cons(x1, zl1))[zl1 / cons(x1, zl1), zr2 / cons(x1, zr2), zr3 / s(zr3)]n[zl1 / x0, zr2 / x0, zr3 / length(x0)] → F(eq(s(s(zr3)), s(length(cons(x1, zr2)))), append(cons(a, nil), cons(x1, zr2)))[zl1 / cons(x1, zl1), zr2 / cons(x1, zr2), zr3 / s(zr3)]n[zl1 / x0, zr2 / x0, zr3 / length(x0)]
by Narrowing at position: [0,1]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
F(true, cons(y1, zs1))[zs1 / cons(y1, zs1), zt1 / s(zt1)]n[zs1 / y0, zt1 / length(y0)] → F(eq(s(s(zt1)), length(cons(a, cons(y1, zs1)))), append(cons(a, nil), cons(y1, zs1)))[zs1 / cons(y1, zs1), zt1 / s(zt1)]n[zs1 / y0, zt1 / length(y0)]
by Narrowing at position: [0,0,0]
intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Instantiate Sigma - Instantiation - Instantiation
F(true, xs)[ ]n[ ] → F(eq(s(length(xs)), length(cons(a, xs))), append(cons(a, nil), xs))[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
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
length(cons(x, xs))[ ]n[ ] → s(length(xs))[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Instantiate Sigma - Instantiation - Instantiation - Instantiation
length(cons(x, xs))[ ]n[ ] → s(length(xs))[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
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
length(cons(x, xs))[ ]n[ ] → s(length(xs))[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Expand Sigma - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
eq(s(x), s(y))[x / s(x), y / s(y)]n[ ] → eq(x, y)[ ]n[ ]
by PatternCreation I
eq(s(x), s(y))[ ]n[ ] → eq(x, y)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused)
length(nil)[ ]n[ ] → 0[ ]n[ ]
by OriginalRule from TRS R
(29) NO