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, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
DIV2(s1(x), s1(y)) -> LT2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
DIV2(s1(x), s1(y)) -> -12(x, y)
LT2(s1(x), s1(y)) -> LT2(x, y)
DIV2(s1(x), s1(y)) -> IF3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
DIV2(s1(x), s1(y)) -> DIV2(-2(x, y), s1(y))
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
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:
DIV2(s1(x), s1(y)) -> LT2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
DIV2(s1(x), s1(y)) -> -12(x, y)
LT2(s1(x), s1(y)) -> LT2(x, y)
DIV2(s1(x), s1(y)) -> IF3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
DIV2(s1(x), s1(y)) -> DIV2(-2(x, y), s1(y))
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LT2(s1(x), s1(y)) -> LT2(x, y)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
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.
LT2(s1(x), s1(y)) -> LT2(x, y)
Used argument filtering: LT2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ 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, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
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
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, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
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
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
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
Q DP problem:
The TRS P consists of the following rules:
DIV2(s1(x), s1(y)) -> DIV2(-2(x, y), s1(y))
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
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.
DIV2(s1(x), s1(y)) -> DIV2(-2(x, y), s1(y))
Used argument filtering: DIV2(x1, x2) = x1
s1(x1) = s1(x1)
-2(x1, x2) = x1
0 = 0
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ 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, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
lt2(x, 0) -> false
lt2(0, s1(y)) -> true
lt2(s1(x), s1(y)) -> lt2(x, y)
if3(true, x, y) -> x
if3(false, x, y) -> y
div2(x, 0) -> 0
div2(0, y) -> 0
div2(s1(x), s1(y)) -> if3(lt2(x, y), 0, s1(div2(-2(x, y), s1(y))))
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.