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:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
DIV2(div2(x, y), z) -> TIMES2(y, z)
PLUS2(s1(x), y) -> PLUS2(x, y)
TIMES2(s1(x), y) -> PLUS2(y, times2(x, y))
QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
TIMES2(s1(x), y) -> TIMES2(x, y)
DIV2(x, y) -> QUOT3(x, y, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(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:
DIV2(div2(x, y), z) -> TIMES2(y, z)
PLUS2(s1(x), y) -> PLUS2(x, y)
TIMES2(s1(x), y) -> PLUS2(y, times2(x, y))
QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
TIMES2(s1(x), y) -> TIMES2(x, y)
DIV2(x, y) -> QUOT3(x, y, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS2(s1(x), y) -> PLUS2(x, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(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.
PLUS2(s1(x), y) -> PLUS2(x, y)
Used argument filtering: PLUS2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: 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:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(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
Q DP problem:
The TRS P consists of the following rules:
TIMES2(s1(x), y) -> TIMES2(x, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(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.
TIMES2(s1(x), y) -> TIMES2(x, y)
Used argument filtering: TIMES2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: 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:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(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
Q DP problem:
The TRS P consists of the following rules:
QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
DIV2(x, y) -> QUOT3(x, y, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(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.
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
Used argument filtering: QUOT3(x1, x2, x3) = x1
DIV2(x1, x2) = x1
s1(x1) = s1(x1)
div2(x1, x2) = x1
times2(x1, x2) = times2(x1, x2)
0 = 0
plus2(x1, x2) = plus2(x1, x2)
Used ordering: Precedence:
times2 > plus2 > s1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
DIV2(x, y) -> QUOT3(x, y, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(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.
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
Used argument filtering: QUOT3(x1, x2, x3) = x1
DIV2(x1, x2) = x1
div2(x1, x2) = div1(x1)
times2(x1, x2) = times2(x1, x2)
0 = 0
s1(x1) = s1(x1)
plus2(x1, x2) = plus2(x1, x2)
Used ordering: Precedence:
times2 > plus2 > s1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
DIV2(x, y) -> QUOT3(x, y, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(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.
QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
Used argument filtering: QUOT3(x1, x2, x3) = x2
0 = 0
DIV2(x1, x2) = x2
s1(x1) = s
Used ordering: Precedence:
0 > s
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
DIV2(x, y) -> QUOT3(x, y, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.