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(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
PLUS2(s1(X), Y) -> PLUS2(X, Y)
MIN2(min2(X, Y), Z) -> PLUS2(Y, Z)
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
QUOT2(s1(X), s1(Y)) -> MIN2(X, Y)
MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
The TRS R consists of the following rules:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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:
PLUS2(s1(X), Y) -> PLUS2(X, Y)
MIN2(min2(X, Y), Z) -> PLUS2(Y, Z)
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
QUOT2(s1(X), s1(Y)) -> MIN2(X, Y)
MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
The TRS R consists of the following rules:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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 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(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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.
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:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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:
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))
The TRS R consists of the following rules:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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.
MIN2(min2(X, Y), Z) -> MIN2(X, plus2(Y, Z))
Used argument filtering: MIN2(x1, x2) = x1
s1(x1) = x1
min2(x1, x2) = min1(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:
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
The TRS R consists of the following rules:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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.
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
Used argument filtering: MIN2(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:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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:
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
The TRS R consists of the following rules:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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.
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
Used argument filtering: QUOT2(x1, x2) = x1
s1(x1) = s1(x1)
min2(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:
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
min2(min2(X, Y), Z) -> min2(X, plus2(Y, Z))
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(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.