(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(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:
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))
A__2ND(cons(X, X1)) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(2nd(X)) → A__2ND(mark(X))
MARK(2nd(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
The TRS R consists of the following rules:
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(cons1(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
mark(from(X)) → a__from(mark(X))
mark(2nd(X)) → a__2nd(mark(X))
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__2nd(X) → 2nd(X)
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__from(X) → from(X)
a__from(X) → cons(mark(X), from(s(X)))
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))
A__2ND(cons(X, X1)) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(2nd(X)) → A__2ND(mark(X))
MARK(2nd(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
The TRS R consists of the following rules:
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(2nd(x1)) = x1
POL(A__2ND(x1)) = x1
POL(A__FROM(x1)) = x1
POL(MARK(x1)) = x1
POL(a__2nd(x1)) = x1
POL(a__from(x1)) = 1 + x1
POL(cons(x1, x2)) = x1 + x2
POL(cons1(x1, x2)) = x1 + x2
POL(from(x1)) = 1 + x1
POL(mark(x1)) = x1
POL(s(x1)) = 0
The following usable rules [FROCOS05] were oriented:
mark(from(X)) → a__from(mark(X))
mark(2nd(X)) → a__2nd(mark(X))
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__2nd(X) → 2nd(X)
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__from(X) → from(X)
a__from(X) → cons(mark(X), from(s(X)))
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))
A__2ND(cons(X, X1)) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(2nd(X)) → A__2ND(mark(X))
MARK(2nd(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
The TRS R consists of the following rules:
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(2nd(X)) → A__2ND(mark(X))
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
MARK(2nd(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))
A__2ND(cons(X, X1)) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X1)
The TRS R consists of the following rules:
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(2nd(X)) → A__2ND(mark(X))
MARK(2nd(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(2nd(x1)) = 1 + x1
POL(A__2ND(x1)) = x1
POL(MARK(x1)) = x1
POL(a__2nd(x1)) = 1 + x1
POL(a__from(x1)) = x1
POL(cons(x1, x2)) = x1 + x2
POL(cons1(x1, x2)) = x1 + x2
POL(from(x1)) = x1
POL(mark(x1)) = x1
POL(s(x1)) = 0
The following usable rules [FROCOS05] were oriented:
mark(from(X)) → a__from(mark(X))
mark(2nd(X)) → a__2nd(mark(X))
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__2nd(X) → 2nd(X)
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__from(X) → from(X)
a__from(X) → cons(mark(X), from(s(X)))
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A__2ND(cons1(X, cons(Y, Z))) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
A__2ND(cons(X, X1)) → A__2ND(cons1(mark(X), mark(X1)))
A__2ND(cons(X, X1)) → MARK(X)
A__2ND(cons(X, X1)) → MARK(X1)
The TRS R consists of the following rules:
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
The TRS R consists of the following rules:
a__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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:
- MARK(cons1(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
- MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
- MARK(cons1(X1, X2)) → MARK(X2)
The graph contains the following edges 1 > 1
(16) TRUE