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:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
PLUS2(s1(x), y) -> PLUS2(x, y)
MINUS2(minus2(x, y), z) -> PLUS2(y, z)
MINUS2(minus2(x, y), z) -> MINUS2(x, plus2(y, z))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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:
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
PLUS2(s1(x), y) -> PLUS2(x, y)
MINUS2(minus2(x, y), z) -> PLUS2(y, z)
MINUS2(minus2(x, y), z) -> MINUS2(x, plus2(y, z))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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: 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:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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:
MINUS2(minus2(x, y), z) -> MINUS2(x, plus2(y, z))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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.
MINUS2(minus2(x, y), z) -> MINUS2(x, plus2(y, z))
Used argument filtering: MINUS2(x1, x2) = x1
minus2(x1, x2) = minus1(x1)
s1(x1) = x1
plus2(x1, x2) = x2
0 = 0
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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.
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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:
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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.
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
Used argument filtering: QUOT2(x1, x2) = x1
s1(x1) = s1(x1)
minus2(x1, x2) = x1
plus2(x1, x2) = plus2(x1, x2)
0 = 0
Used ordering: Quasi Precedence:
plus_2 > s_1
↳ 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:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
minus2(minus2(x, y), z) -> minus2(x, plus2(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.