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:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F3(x, s1(y), z) -> ODD1(s1(y))
POW2(x, y) -> F3(x, y, s1(0))
F3(x, s1(y), z) -> *12(x, x)
F3(x, s1(y), z) -> F3(*2(x, x), half1(s1(y)), z)
*12(x, s1(y)) -> *12(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
F3(x, s1(y), z) -> F3(x, y, *2(x, z))
F3(x, s1(y), z) -> HALF1(s1(y))
F3(x, s1(y), z) -> *12(x, z)
ODD1(s1(s1(x))) -> ODD1(x)
F3(x, s1(y), z) -> IF3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
HALF1(s1(s1(x))) -> HALF1(x)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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:
F3(x, s1(y), z) -> ODD1(s1(y))
POW2(x, y) -> F3(x, y, s1(0))
F3(x, s1(y), z) -> *12(x, x)
F3(x, s1(y), z) -> F3(*2(x, x), half1(s1(y)), z)
*12(x, s1(y)) -> *12(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
F3(x, s1(y), z) -> F3(x, y, *2(x, z))
F3(x, s1(y), z) -> HALF1(s1(y))
F3(x, s1(y), z) -> *12(x, z)
ODD1(s1(s1(x))) -> ODD1(x)
F3(x, s1(y), z) -> IF3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
HALF1(s1(s1(x))) -> HALF1(x)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HALF1(s1(s1(x))) -> HALF1(x)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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.
HALF1(s1(s1(x))) -> HALF1(x)
Used argument filtering: HALF1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ODD1(s1(s1(x))) -> ODD1(x)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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.
ODD1(s1(s1(x))) -> ODD1(x)
Used argument filtering: ODD1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
*12(x, s1(y)) -> *12(x, y)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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(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
↳ 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(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F3(x, s1(y), z) -> F3(*2(x, x), half1(s1(y)), z)
F3(x, s1(y), z) -> F3(x, y, *2(x, z))
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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.
F3(x, s1(y), z) -> F3(x, y, *2(x, z))
Used argument filtering: F3(x1, x2, x3) = x2
s1(x1) = s1(x1)
half1(x1) = x1
*2(x1, x2) = *
0 = 0
+2(x1, x2) = +
Used ordering: Quasi Precedence:
* > 0
* > +
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F3(x, s1(y), z) -> F3(*2(x, x), half1(s1(y)), z)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
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(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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
↳ 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(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
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.