Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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))
+12(s1(x), s1(y)) -> +12(x, y)
PROD1(cons2(x, l)) -> *12(x, prod1(l))
*12(*2(x, y), z) -> *12(y, z)
PROD1(cons2(x, l)) -> PROD1(l)
+12(+2(x, y), z) -> +12(y, z)
*12(s1(x), s1(y)) -> +12(*2(x, y), +2(x, y))
*12(s1(x), s1(y)) -> +12(x, y)
*12(s1(x), s1(y)) -> *12(x, y)
SUM1(cons2(x, l)) -> SUM1(l)
*12(*2(x, y), z) -> *12(x, *2(y, z))
+12(+2(x, y), z) -> +12(x, +2(y, z))
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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))
+12(s1(x), s1(y)) -> +12(x, y)
PROD1(cons2(x, l)) -> *12(x, prod1(l))
*12(*2(x, y), z) -> *12(y, z)
PROD1(cons2(x, l)) -> PROD1(l)
+12(+2(x, y), z) -> +12(y, z)
*12(s1(x), s1(y)) -> +12(*2(x, y), +2(x, y))
*12(s1(x), s1(y)) -> +12(x, y)
*12(s1(x), s1(y)) -> *12(x, y)
SUM1(cons2(x, l)) -> SUM1(l)
*12(*2(x, y), z) -> *12(x, *2(y, z))
+12(+2(x, y), z) -> +12(x, +2(y, z))
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(s1(x), s1(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
Used argument filtering: +12(x1, x2) = x1
s1(x1) = x1
+2(x1, x2) = +2(x1, x2)
0 = 0
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(s1(x), s1(y)) -> +12(x, y)
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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(s1(x), s1(y)) -> +12(x, y)
Used argument filtering: +12(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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
↳ 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:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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(*2(x, y), z) -> *12(y, z)
*12(s1(x), s1(y)) -> *12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, z))
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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(*2(x, y), z) -> *12(y, z)
*12(s1(x), s1(y)) -> *12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, z))
Used argument filtering: *12(x1, x2) = x1
*2(x1, x2) = *2(x1, x2)
s1(x1) = s1(x1)
0 = 0
+2(x1, x2) = +2(x1, x2)
Used ordering: Quasi Precedence:
*_2 > +_2 > s_1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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:
+2(x, 0) -> x
+2(0, x) -> x
+2(s1(x), s1(y)) -> s1(s1(+2(x, y)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(x, 0) -> 0
*2(0, x) -> 0
*2(s1(x), s1(y)) -> s1(+2(*2(x, y), +2(x, y)))
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 0
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> s1(0)
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.