Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
P1(p1(s1(x))) -> P1(x)
MINUS2(x, y) -> LE2(x, y)
MINUS2(x, y) -> IF3(le2(x, y), x, y)
LE2(p1(s1(x)), x) -> LE2(x, x)
LE2(s1(x), s1(y)) -> LE2(x, y)
IF3(false, x, y) -> MINUS2(p1(x), y)
IF3(false, x, y) -> P1(x)
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(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:
P1(p1(s1(x))) -> P1(x)
MINUS2(x, y) -> LE2(x, y)
MINUS2(x, y) -> IF3(le2(x, y), x, y)
LE2(p1(s1(x)), x) -> LE2(x, x)
LE2(s1(x), s1(y)) -> LE2(x, y)
IF3(false, x, y) -> MINUS2(p1(x), y)
IF3(false, x, y) -> P1(x)
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), 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:
LE2(p1(s1(x)), x) -> LE2(x, x)
LE2(s1(x), s1(y)) -> LE2(x, y)
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(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.
LE2(s1(x), s1(y)) -> LE2(x, y)
Used argument filtering: LE2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LE2(p1(s1(x)), x) -> LE2(x, x)
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(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.
LE2(p1(s1(x)), x) -> LE2(x, x)
Used argument filtering: LE2(x1, x2) = x1
p1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P1(p1(s1(x))) -> P1(x)
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(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.
P1(p1(s1(x))) -> P1(x)
Used argument filtering: P1(x1) = x1
p1(x1) = x1
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:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS2(x, y) -> IF3(le2(x, y), x, y)
IF3(false, x, y) -> MINUS2(p1(x), y)
The TRS R consists of the following rules:
p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.