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:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
SUM1(cons2(x, l)) -> +12(x, sum1(l))
SUM1(nil) -> 011(#)
+12(11(x), 11(y)) -> +12(x, y)
SUM1(cons2(x, l)) -> SUM1(l)
+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
PROD1(cons2(x, l)) -> *12(x, prod1(l))
PROD1(cons2(x, l)) -> PROD1(l)
*12(11(x), y) -> 011(*2(x, y))
*12(01(x), y) -> 011(*2(x, y))
*12(11(x), y) -> *12(x, y)
+12(01(x), 01(y)) -> 011(+2(x, y))
*12(01(x), y) -> *12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
*12(11(x), y) -> +12(01(*2(x, y)), y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
SUM1(cons2(x, l)) -> +12(x, sum1(l))
SUM1(nil) -> 011(#)
+12(11(x), 11(y)) -> +12(x, y)
SUM1(cons2(x, l)) -> SUM1(l)
+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
PROD1(cons2(x, l)) -> *12(x, prod1(l))
PROD1(cons2(x, l)) -> PROD1(l)
*12(11(x), y) -> 011(*2(x, y))
*12(01(x), y) -> 011(*2(x, y))
*12(11(x), y) -> *12(x, y)
+12(01(x), 01(y)) -> 011(+2(x, y))
*12(01(x), y) -> *12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
*12(11(x), y) -> +12(01(*2(x, y)), y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 8 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SUM1(cons2(x, l)) -> SUM1(l)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
SUM1(cons2(x, l)) -> SUM1(l)
Used argument filtering: SUM1(x1) = x1
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
*12(11(x), y) -> *12(x, y)
*12(01(x), y) -> *12(x, y)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
*12(01(x), y) -> *12(x, y)
Used argument filtering: *12(x1, x2) = x1
11(x1) = x1
01(x1) = 01(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
*12(11(x), y) -> *12(x, y)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
*12(11(x), y) -> *12(x, y)
Used argument filtering: *12(x1, x2) = x1
11(x1) = 11(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
PROD1(cons2(x, l)) -> PROD1(l)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
PROD1(cons2(x, l)) -> PROD1(l)
Used argument filtering: PROD1(x1) = x1
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.