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:
sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1
Q is empty.
We have reversed the following QTRS:
The set of rules R is
sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1
The set Q is empty.
We have obtained the following QTRS:
0(sq(x)) → s(p(s(p(0(s(s(s(s(p(p(p(p(s(p(s(p(x)))))))))))))))))
s(sq(x)) → s(s(s(s(s(s(p(p(p(p(p(p(sq(s(s(s(p(p(p(s(p(s(p(twice(s(s(p(p(s(p(s(p(s(x)))))))))))))))))))))))))))))))))
0(twice(x)) → s(p(s(p(s(p(s(s(p(p(s(s(s(p(p(p(0(s(s(s(p(s(s(p(p(p(p(x)))))))))))))))))))))))))))
s(twice(x)) → s(p(s(p(twice(s(s(s(p(p(s(s(s(p(p(x)))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(s(s(s(s(s(s(s(0(x))))))))))))
0(x) → x
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
0(sq(x)) → s(p(s(p(0(s(s(s(s(p(p(p(p(s(p(s(p(x)))))))))))))))))
s(sq(x)) → s(s(s(s(s(s(p(p(p(p(p(p(sq(s(s(s(p(p(p(s(p(s(p(twice(s(s(p(p(s(p(s(p(s(x)))))))))))))))))))))))))))))))))
0(twice(x)) → s(p(s(p(s(p(s(s(p(p(s(s(s(p(p(p(0(s(s(s(p(s(s(p(p(p(p(x)))))))))))))))))))))))))))
s(twice(x)) → s(p(s(p(twice(s(s(s(p(p(s(s(s(p(p(x)))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(s(s(s(s(s(s(s(0(x))))))))))))
0(x) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1
The set Q is empty.
We have obtained the following QTRS:
0(sq(x)) → s(p(s(p(0(s(s(s(s(p(p(p(p(s(p(s(p(x)))))))))))))))))
s(sq(x)) → s(s(s(s(s(s(p(p(p(p(p(p(sq(s(s(s(p(p(p(s(p(s(p(twice(s(s(p(p(s(p(s(p(s(x)))))))))))))))))))))))))))))))))
0(twice(x)) → s(p(s(p(s(p(s(s(p(p(s(s(s(p(p(p(0(s(s(s(p(s(s(p(p(p(p(x)))))))))))))))))))))))))))
s(twice(x)) → s(p(s(p(twice(s(s(s(p(p(s(s(s(p(p(x)))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(s(s(s(s(s(s(s(0(x))))))))))))
0(x) → x
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
0(sq(x)) → s(p(s(p(0(s(s(s(s(p(p(p(p(s(p(s(p(x)))))))))))))))))
s(sq(x)) → s(s(s(s(s(s(p(p(p(p(p(p(sq(s(s(s(p(p(p(s(p(s(p(twice(s(s(p(p(s(p(s(p(s(x)))))))))))))))))))))))))))))))))
0(twice(x)) → s(p(s(p(s(p(s(s(p(p(s(s(s(p(p(p(0(s(s(s(p(s(s(p(p(p(p(x)))))))))))))))))))))))))))
s(twice(x)) → s(p(s(p(twice(s(s(s(p(p(s(s(s(p(p(x)))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(s(s(s(s(s(s(s(0(x))))))))))))
0(x) → x
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
Used ordering:
Polynomial interpretation [25]:
POL(0(x1)) = 2·x1
POL(p(x1)) = x1
POL(s(x1)) = x1
POL(sq(x1)) = 2 + 2·x1
POL(twice(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
0(x1) → x1
Used ordering:
Polynomial interpretation [25]:
POL(0(x1)) = 2 + 2·x1
POL(p(x1)) = x1
POL(s(x1)) = x1
POL(sq(x1)) = 2·x1
POL(twice(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
TWICE(0(x1)) → P(p(s(s(p(s(p(s(p(s(x1))))))))))
TWICE(0(x1)) → P(s(p(s(p(s(x1))))))
SQ(s(x1)) → P(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(s(s(s(x1)))))))))
TWICE(0(x1)) → P(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))
SQ(s(x1)) → TWICE(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))
TWICE(0(x1)) → P(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))
SQ(s(x1)) → P(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(twice(p(s(p(s(x1)))))))))
TWICE(0(x1)) → P(s(s(p(s(p(s(p(s(x1)))))))))
SQ(s(x1)) → P(p(p(p(p(s(s(s(s(s(s(x1)))))))))))
TWICE(0(x1)) → P(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))
TWICE(s(x1)) → P(s(x1))
P(p(s(x1))) → P(x1)
TWICE(0(x1)) → P(s(x1))
SQ(s(x1)) → P(p(s(s(s(s(s(s(x1))))))))
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
SQ(s(x1)) → P(p(p(p(s(s(s(s(s(s(x1))))))))))
TWICE(s(x1)) → P(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
TWICE(0(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(s(s(s(s(s(x1)))))))
TWICE(0(x1)) → P(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))
TWICE(0(x1)) → P(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))
SQ(s(x1)) → P(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
SQ(s(x1)) → P(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))
TWICE(s(x1)) → P(p(s(s(s(twice(p(s(p(s(x1))))))))))
TWICE(s(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))
SQ(s(x1)) → P(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))
SQ(s(x1)) → P(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))
TWICE(0(x1)) → P(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
TWICE(0(x1)) → P(p(s(s(p(s(p(s(p(s(x1))))))))))
TWICE(0(x1)) → P(s(p(s(p(s(x1))))))
SQ(s(x1)) → P(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(s(s(s(x1)))))))))
TWICE(0(x1)) → P(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))
SQ(s(x1)) → TWICE(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))
TWICE(0(x1)) → P(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))
SQ(s(x1)) → P(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(twice(p(s(p(s(x1)))))))))
TWICE(0(x1)) → P(s(s(p(s(p(s(p(s(x1)))))))))
SQ(s(x1)) → P(p(p(p(p(s(s(s(s(s(s(x1)))))))))))
TWICE(0(x1)) → P(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))
TWICE(s(x1)) → P(s(x1))
P(p(s(x1))) → P(x1)
TWICE(0(x1)) → P(s(x1))
SQ(s(x1)) → P(p(s(s(s(s(s(s(x1))))))))
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
SQ(s(x1)) → P(p(p(p(s(s(s(s(s(s(x1))))))))))
TWICE(s(x1)) → P(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
TWICE(0(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(s(s(s(s(s(x1)))))))
TWICE(0(x1)) → P(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))
TWICE(0(x1)) → P(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))
SQ(s(x1)) → P(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
SQ(s(x1)) → P(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))
TWICE(s(x1)) → P(p(s(s(s(twice(p(s(p(s(x1))))))))))
TWICE(s(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))
SQ(s(x1)) → P(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))
SQ(s(x1)) → P(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))
TWICE(0(x1)) → P(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 35 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x1))) → P(x1)
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x1))) → P(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x1))) → P(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
P(p(s(x1))) → P(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(P(x1)) = 2·x1
POL(p(x1)) = 2·x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
R is empty.
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
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
sq(s(x0))
twice(0(x0))
twice(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QReductionProof
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as they contain symbols which do neither occur in P nor in R.
p(0(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
We have to consider all (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
Q is empty.
We have to consider all (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
sq(s(x0))
twice(0(x0))
twice(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule TWICE(s(x1)) → TWICE(p(s(p(s(x1))))) at position [0] we obtained the following new rules:
TWICE(s(x1)) → TWICE(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule TWICE(s(x1)) → TWICE(p(s(x1))) at position [0] we obtained the following new rules:
TWICE(s(x1)) → TWICE(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(x1)
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(x1)
R is empty.
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
p(s(x0))
p(0(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE(s(x1)) → TWICE(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
TWICE(s(x1)) → TWICE(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(TWICE(x1)) = 2·x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
R is empty.
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
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
sq(s(x0))
twice(0(x0))
twice(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))) at position [0,0,0,0,0,0] we obtained the following new rules:
SQ(s(x1)) → SQ(p(p(p(p(p(s(s(s(s(s(x1)))))))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(s(s(s(s(s(x1)))))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(p(p(p(s(s(s(s(s(x1))))))))))) at position [0,0,0,0,0] we obtained the following new rules:
SQ(s(x1)) → SQ(p(p(p(p(s(s(s(s(x1)))))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(s(s(s(s(x1)))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(p(p(s(s(s(s(x1))))))))) at position [0,0,0,0] we obtained the following new rules:
SQ(s(x1)) → SQ(p(p(p(s(s(s(x1)))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(s(s(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(p(s(s(s(x1))))))) at position [0,0,0] we obtained the following new rules:
SQ(s(x1)) → SQ(p(p(s(s(x1)))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(s(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(s(s(x1))))) at position [0,0] we obtained the following new rules:
SQ(s(x1)) → SQ(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(s(x1))) at position [0] we obtained the following new rules:
SQ(s(x1)) → SQ(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(x1)
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(x1)
R is empty.
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
p(s(x0))
p(0(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
SQ(s(x1)) → SQ(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(SQ(x1)) = 2·x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ MNOCProof
Q DP problem:
P is empty.
R is empty.
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.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
The TRS R consists of the following rules:
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
sq(s(x0))
twice(0(x0))
twice(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
Q is empty.
We have to consider all (P,Q,R)-chains.