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:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, s1(z))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(s1(z), 0)
TIMES2(x, s1(y)) -> TIMES2(x, y)
TIMES2(x, s1(y)) -> PLUS2(times2(x, y), x)
TIMES2(x, plus2(y, s1(z))) -> PLUS2(y, times2(s1(z), 0))
PLUS2(x, s1(y)) -> PLUS2(x, y)
TIMES2(x, plus2(y, s1(z))) -> PLUS2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
The TRS R consists of the following rules:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, 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:
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, s1(z))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(s1(z), 0)
TIMES2(x, s1(y)) -> TIMES2(x, y)
TIMES2(x, s1(y)) -> PLUS2(times2(x, y), x)
TIMES2(x, plus2(y, s1(z))) -> PLUS2(y, times2(s1(z), 0))
PLUS2(x, s1(y)) -> PLUS2(x, y)
TIMES2(x, plus2(y, s1(z))) -> PLUS2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
The TRS R consists of the following rules:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS2(x, s1(y)) -> PLUS2(x, y)
The TRS R consists of the following rules:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, 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(x, s1(y)) -> PLUS2(x, y)
Used argument filtering: PLUS2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, 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
Q DP problem:
The TRS P consists of the following rules:
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, s1(z))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, s1(y)) -> TIMES2(x, y)
The TRS R consists of the following rules:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, 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.
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, s1(z))
Used argument filtering: TIMES2(x1, x2) = x2
plus2(x1, x2) = plus2(x1, x2)
s1(x1) = x1
times2(x1, x2) = times
0 = 0
Used ordering: Quasi Precedence:
plus_2 > [times, 0]
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, s1(y)) -> TIMES2(x, y)
The TRS R consists of the following rules:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, 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.
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, s1(y)) -> TIMES2(x, y)
Used argument filtering: TIMES2(x1, x2) = x2
plus2(x1, x2) = plus2(x1, x2)
s1(x1) = s1(x1)
times2(x1, x2) = times
0 = 0
Used ordering: Quasi Precedence:
plus_2 > [s_1, times, 0]
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, 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.