Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
MARK(sel(X1, X2)) → MARK(X1)
MARK(from(X)) → MARK(X)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
A__SEL(0, cons(X, Y)) → MARK(X)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(sel(X1, X2)) → MARK(X1)
MARK(from(X)) → MARK(X)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
A__SEL(0, cons(X, Y)) → MARK(X)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
The remaining pairs can at least be oriented weakly.
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
A__SEL(0, cons(X, Y)) → MARK(X)
Used ordering: Polynomial interpretation with max and min functions [25]:
POL(0) = 0
POL(A__FROM(x1)) = x1
POL(A__SEL(x1, x2)) = x1 + x2
POL(MARK(x1)) = x1
POL(a__from(x1)) = 1 + x1
POL(a__sel(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = max(x1, x2)
POL(from(x1)) = 1 + x1
POL(mark(x1)) = x1
POL(s(x1)) = x1
POL(sel(x1, x2)) = x1 + x2
The following usable rules [17] were oriented:
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MARK(sel(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
A__SEL(0, cons(X, Y)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(sel(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(0, cons(X, Y)) → MARK(X)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
A__SEL(0, cons(X, Y)) → MARK(X)
The remaining pairs can at least be oriented weakly.
MARK(sel(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
Used ordering: Polynomial interpretation with max and min functions [25]:
POL(0) = 1
POL(A__SEL(x1, x2)) = x1 + x2
POL(MARK(x1)) = x1
POL(a__from(x1)) = x1
POL(a__sel(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = max(x1, x2)
POL(from(x1)) = x1
POL(mark(x1)) = x1
POL(s(x1)) = x1
POL(sel(x1, x2)) = x1 + x2
The following usable rules [17] were oriented:
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
MARK(sel(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
Used ordering: Polynomial interpretation with max and min functions [25]:
POL(0) = 0
POL(A__SEL(x1, x2)) = x1 + x2
POL(MARK(x1)) = x1
POL(a__from(x1)) = x1
POL(a__sel(x1, x2)) = 1 + x1 + x2
POL(cons(x1, x2)) = max(x1, x2)
POL(from(x1)) = x1
POL(mark(x1)) = x1
POL(s(x1)) = x1
POL(sel(x1, x2)) = 1 + x1 + x2
The following usable rules [17] were oriented:
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(s(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(cons(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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(s(X)) → MARK(X)
The graph contains the following edges 1 > 1
- MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z)) at position [0] we obtained the following new rules:
A__SEL(s(from(x0)), cons(y1, y2)) → A__SEL(a__from(mark(x0)), mark(y2))
A__SEL(s(0), cons(y1, y2)) → A__SEL(0, mark(y2))
A__SEL(s(sel(x0, x1)), cons(y1, y2)) → A__SEL(a__sel(mark(x0), mark(x1)), mark(y2))
A__SEL(s(s(x0)), cons(y1, y2)) → A__SEL(s(mark(x0)), mark(y2))
A__SEL(s(cons(x0, x1)), cons(y1, y2)) → A__SEL(cons(mark(x0), x1), mark(y2))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(from(x0)), cons(y1, y2)) → A__SEL(a__from(mark(x0)), mark(y2))
A__SEL(s(sel(x0, x1)), cons(y1, y2)) → A__SEL(a__sel(mark(x0), mark(x1)), mark(y2))
A__SEL(s(0), cons(y1, y2)) → A__SEL(0, mark(y2))
A__SEL(s(s(x0)), cons(y1, y2)) → A__SEL(s(mark(x0)), mark(y2))
A__SEL(s(cons(x0, x1)), cons(y1, y2)) → A__SEL(cons(mark(x0), x1), mark(y2))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(from(x0)), cons(y1, y2)) → A__SEL(a__from(mark(x0)), mark(y2))
A__SEL(s(sel(x0, x1)), cons(y1, y2)) → A__SEL(a__sel(mark(x0), mark(x1)), mark(y2))
A__SEL(s(s(x0)), cons(y1, y2)) → A__SEL(s(mark(x0)), mark(y2))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__SEL(s(from(x0)), cons(y1, y2)) → A__SEL(a__from(mark(x0)), mark(y2)) at position [1] we obtained the following new rules:
A__SEL(s(from(y0)), cons(y1, cons(x0, x1))) → A__SEL(a__from(mark(y0)), cons(mark(x0), x1))
A__SEL(s(from(y0)), cons(y1, sel(x0, x1))) → A__SEL(a__from(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(from(y0)), cons(y1, from(x0))) → A__SEL(a__from(mark(y0)), a__from(mark(x0)))
A__SEL(s(from(y0)), cons(y1, 0)) → A__SEL(a__from(mark(y0)), 0)
A__SEL(s(from(y0)), cons(y1, s(x0))) → A__SEL(a__from(mark(y0)), s(mark(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(from(y0)), cons(y1, cons(x0, x1))) → A__SEL(a__from(mark(y0)), cons(mark(x0), x1))
A__SEL(s(sel(x0, x1)), cons(y1, y2)) → A__SEL(a__sel(mark(x0), mark(x1)), mark(y2))
A__SEL(s(s(x0)), cons(y1, y2)) → A__SEL(s(mark(x0)), mark(y2))
A__SEL(s(from(y0)), cons(y1, sel(x0, x1))) → A__SEL(a__from(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(from(y0)), cons(y1, from(x0))) → A__SEL(a__from(mark(y0)), a__from(mark(x0)))
A__SEL(s(from(y0)), cons(y1, 0)) → A__SEL(a__from(mark(y0)), 0)
A__SEL(s(from(y0)), cons(y1, s(x0))) → A__SEL(a__from(mark(y0)), s(mark(x0)))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(from(y0)), cons(y1, cons(x0, x1))) → A__SEL(a__from(mark(y0)), cons(mark(x0), x1))
A__SEL(s(sel(x0, x1)), cons(y1, y2)) → A__SEL(a__sel(mark(x0), mark(x1)), mark(y2))
A__SEL(s(s(x0)), cons(y1, y2)) → A__SEL(s(mark(x0)), mark(y2))
A__SEL(s(from(y0)), cons(y1, sel(x0, x1))) → A__SEL(a__from(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(from(y0)), cons(y1, from(x0))) → A__SEL(a__from(mark(y0)), a__from(mark(x0)))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__SEL(s(sel(x0, x1)), cons(y1, y2)) → A__SEL(a__sel(mark(x0), mark(x1)), mark(y2)) at position [1] we obtained the following new rules:
A__SEL(s(sel(y0, y1)), cons(y2, 0)) → A__SEL(a__sel(mark(y0), mark(y1)), 0)
A__SEL(s(sel(y0, y1)), cons(y2, sel(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), a__sel(mark(x0), mark(x1)))
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(sel(y0, y1)), cons(y2, s(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), s(mark(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(sel(y0, y1)), cons(y2, sel(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), a__sel(mark(x0), mark(x1)))
A__SEL(s(sel(y0, y1)), cons(y2, 0)) → A__SEL(a__sel(mark(y0), mark(y1)), 0)
A__SEL(s(from(y0)), cons(y1, cons(x0, x1))) → A__SEL(a__from(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(x0)), cons(y1, y2)) → A__SEL(s(mark(x0)), mark(y2))
A__SEL(s(from(y0)), cons(y1, sel(x0, x1))) → A__SEL(a__from(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(from(y0)), cons(y1, from(x0))) → A__SEL(a__from(mark(y0)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(sel(y0, y1)), cons(y2, s(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), s(mark(x0)))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(sel(y0, y1)), cons(y2, sel(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), a__sel(mark(x0), mark(x1)))
A__SEL(s(from(y0)), cons(y1, cons(x0, x1))) → A__SEL(a__from(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(x0)), cons(y1, y2)) → A__SEL(s(mark(x0)), mark(y2))
A__SEL(s(from(y0)), cons(y1, sel(x0, x1))) → A__SEL(a__from(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(from(y0)), cons(y1, from(x0))) → A__SEL(a__from(mark(y0)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__SEL(s(s(x0)), cons(y1, y2)) → A__SEL(s(mark(x0)), mark(y2)) at position [1] we obtained the following new rules:
A__SEL(s(s(y0)), cons(y1, s(x0))) → A__SEL(s(mark(y0)), s(mark(x0)))
A__SEL(s(s(y0)), cons(y1, sel(x0, x1))) → A__SEL(s(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(s(y0)), cons(y1, 0)) → A__SEL(s(mark(y0)), 0)
A__SEL(s(s(y0)), cons(y1, cons(x0, x1))) → A__SEL(s(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(s(y0)), cons(y1, s(x0))) → A__SEL(s(mark(y0)), s(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, sel(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), a__sel(mark(x0), mark(x1)))
A__SEL(s(s(y0)), cons(y1, sel(x0, x1))) → A__SEL(s(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(from(y0)), cons(y1, cons(x0, x1))) → A__SEL(a__from(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, 0)) → A__SEL(s(mark(y0)), 0)
A__SEL(s(from(y0)), cons(y1, sel(x0, x1))) → A__SEL(a__from(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(from(y0)), cons(y1, from(x0))) → A__SEL(a__from(mark(y0)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, cons(x0, x1))) → A__SEL(s(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(sel(y0, y1)), cons(y2, sel(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), a__sel(mark(x0), mark(x1)))
A__SEL(s(s(y0)), cons(y1, sel(x0, x1))) → A__SEL(s(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(from(y0)), cons(y1, cons(x0, x1))) → A__SEL(a__from(mark(y0)), cons(mark(x0), x1))
A__SEL(s(from(y0)), cons(y1, sel(x0, x1))) → A__SEL(a__from(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(from(y0)), cons(y1, from(x0))) → A__SEL(a__from(mark(y0)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, cons(x0, x1))) → A__SEL(s(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
A__SEL(s(from(y0)), cons(y1, cons(x0, x1))) → A__SEL(a__from(mark(y0)), cons(mark(x0), x1))
A__SEL(s(from(y0)), cons(y1, sel(x0, x1))) → A__SEL(a__from(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(from(y0)), cons(y1, from(x0))) → A__SEL(a__from(mark(y0)), a__from(mark(x0)))
The remaining pairs can at least be oriented weakly.
A__SEL(s(sel(y0, y1)), cons(y2, sel(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), a__sel(mark(x0), mark(x1)))
A__SEL(s(s(y0)), cons(y1, sel(x0, x1))) → A__SEL(s(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, cons(x0, x1))) → A__SEL(s(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 1
POL(A__SEL(x1, x2)) = x1
POL(a__from(x1)) = 0
POL(a__sel(x1, x2)) = x1
POL(cons(x1, x2)) = 0
POL(from(x1)) = 0
POL(mark(x1)) = 1
POL(s(x1)) = 1
POL(sel(x1, x2)) = x1
The following usable rules [17] were oriented:
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(sel(y0, y1)), cons(y2, sel(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), a__sel(mark(x0), mark(x1)))
A__SEL(s(s(y0)), cons(y1, sel(x0, x1))) → A__SEL(s(mark(y0)), a__sel(mark(x0), mark(x1)))
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, cons(x0, x1))) → A__SEL(s(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
A__SEL(s(sel(y0, y1)), cons(y2, sel(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), a__sel(mark(x0), mark(x1)))
A__SEL(s(s(y0)), cons(y1, sel(x0, x1))) → A__SEL(s(mark(y0)), a__sel(mark(x0), mark(x1)))
The remaining pairs can at least be oriented weakly.
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, cons(x0, x1))) → A__SEL(s(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
Used ordering: Polynomial interpretation with max and min functions [25]:
POL(0) = 0
POL(A__SEL(x1, x2)) = x2
POL(a__from(x1)) = x1
POL(a__sel(x1, x2)) = 1 + x2
POL(cons(x1, x2)) = x2 + max(x1, x2)
POL(from(x1)) = x1
POL(mark(x1)) = x1
POL(s(x1)) = 0
POL(sel(x1, x2)) = 1 + x2
The following usable rules [17] were oriented:
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, cons(x0, x1))) → A__SEL(s(mark(y0)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
A__SEL(s(sel(y0, y1)), cons(y2, cons(x0, x1))) → A__SEL(a__sel(mark(y0), mark(y1)), cons(mark(x0), x1))
A__SEL(s(s(y0)), cons(y1, cons(x0, x1))) → A__SEL(s(mark(y0)), cons(mark(x0), x1))
The remaining pairs can at least be oriented weakly.
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( sel(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( cons(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( a__sel(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( A__SEL(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A__SEL(s(sel(y0, y1)), cons(y2, from(x0))) → A__SEL(a__sel(mark(y0), mark(y1)), a__from(mark(x0)))
A__SEL(s(s(y0)), cons(y1, from(x0))) → A__SEL(s(mark(y0)), a__from(mark(x0)))
The TRS R consists of the following rules:
a__from(X) → cons(mark(X), from(s(X)))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
mark(from(X)) → a__from(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__sel(X1, X2) → sel(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.