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:
thrice(0(x1)) → p(s(p(p(p(s(s(s(0(p(s(p(s(x1)))))))))))))
thrice(s(x1)) → p(p(s(s(half(p(p(s(s(p(s(sixtimes(p(s(p(p(s(s(x1))))))))))))))))))
half(0(x1)) → p(p(s(s(p(s(0(p(s(s(s(s(x1))))))))))))
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(0(x1)) → p(s(p(s(0(s(s(s(s(s(p(s(p(s(x1))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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:
thrice(0(x1)) → p(s(p(p(p(s(s(s(0(p(s(p(s(x1)))))))))))))
thrice(s(x1)) → p(p(s(s(half(p(p(s(s(p(s(sixtimes(p(s(p(p(s(s(x1))))))))))))))))))
half(0(x1)) → p(p(s(s(p(s(0(p(s(s(s(s(x1))))))))))))
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(0(x1)) → p(s(p(s(0(s(s(s(s(s(p(s(p(s(x1))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
0(x1) → x1
Q is empty.
We have reversed the following QTRS:
The set of rules R is
thrice(0(x1)) → p(s(p(p(p(s(s(s(0(p(s(p(s(x1)))))))))))))
thrice(s(x1)) → p(p(s(s(half(p(p(s(s(p(s(sixtimes(p(s(p(p(s(s(x1))))))))))))))))))
half(0(x1)) → p(p(s(s(p(s(0(p(s(s(s(s(x1))))))))))))
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(0(x1)) → p(s(p(s(0(s(s(s(s(s(p(s(p(s(x1))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
0(x1) → x1
The set Q is empty.
We have obtained the following QTRS:
0(thrice(x)) → s(p(s(p(0(s(s(s(p(p(p(s(p(x)))))))))))))
s(thrice(x)) → s(s(p(p(s(p(sixtimes(s(p(s(s(p(p(half(s(s(p(p(x))))))))))))))))))
0(half(x)) → s(s(s(s(p(0(s(p(s(s(p(p(x))))))))))))
s(half(x)) → s(p(s(s(p(p(half(s(s(p(p(s(s(p(p(s(p(x)))))))))))))))))
s(s(half(x))) → s(p(s(s(p(p(half(s(s(p(p(s(s(p(s(p(x))))))))))))))))
0(sixtimes(x)) → s(p(s(p(s(s(s(s(s(0(s(p(s(p(x))))))))))))))
s(sixtimes(x)) → s(s(s(p(p(p(s(p(sixtimes(s(s(s(p(s(p(p(s(s(s(s(s(s(s(p(p(x)))))))))))))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → 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(thrice(x)) → s(p(s(p(0(s(s(s(p(p(p(s(p(x)))))))))))))
s(thrice(x)) → s(s(p(p(s(p(sixtimes(s(p(s(s(p(p(half(s(s(p(p(x))))))))))))))))))
0(half(x)) → s(s(s(s(p(0(s(p(s(s(p(p(x))))))))))))
s(half(x)) → s(p(s(s(p(p(half(s(s(p(p(s(s(p(p(s(p(x)))))))))))))))))
s(s(half(x))) → s(p(s(s(p(p(half(s(s(p(p(s(s(p(s(p(x))))))))))))))))
0(sixtimes(x)) → s(p(s(p(s(s(s(s(s(0(s(p(s(p(x))))))))))))))
s(sixtimes(x)) → s(s(s(p(p(p(s(p(sixtimes(s(s(s(p(s(p(p(s(s(s(s(s(s(s(p(p(x)))))))))))))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(0(x)))))
0(x) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
thrice(0(x1)) → p(s(p(p(p(s(s(s(0(p(s(p(s(x1)))))))))))))
thrice(s(x1)) → p(p(s(s(half(p(p(s(s(p(s(sixtimes(p(s(p(p(s(s(x1))))))))))))))))))
half(0(x1)) → p(p(s(s(p(s(0(p(s(s(s(s(x1))))))))))))
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(0(x1)) → p(s(p(s(0(s(s(s(s(s(p(s(p(s(x1))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
0(x1) → x1
The set Q is empty.
We have obtained the following QTRS:
0(thrice(x)) → s(p(s(p(0(s(s(s(p(p(p(s(p(x)))))))))))))
s(thrice(x)) → s(s(p(p(s(p(sixtimes(s(p(s(s(p(p(half(s(s(p(p(x))))))))))))))))))
0(half(x)) → s(s(s(s(p(0(s(p(s(s(p(p(x))))))))))))
s(half(x)) → s(p(s(s(p(p(half(s(s(p(p(s(s(p(p(s(p(x)))))))))))))))))
s(s(half(x))) → s(p(s(s(p(p(half(s(s(p(p(s(s(p(s(p(x))))))))))))))))
0(sixtimes(x)) → s(p(s(p(s(s(s(s(s(0(s(p(s(p(x))))))))))))))
s(sixtimes(x)) → s(s(s(p(p(p(s(p(sixtimes(s(s(s(p(s(p(p(s(s(s(s(s(s(s(p(p(x)))))))))))))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → 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(thrice(x)) → s(p(s(p(0(s(s(s(p(p(p(s(p(x)))))))))))))
s(thrice(x)) → s(s(p(p(s(p(sixtimes(s(p(s(s(p(p(half(s(s(p(p(x))))))))))))))))))
0(half(x)) → s(s(s(s(p(0(s(p(s(s(p(p(x))))))))))))
s(half(x)) → s(p(s(s(p(p(half(s(s(p(p(s(s(p(p(s(p(x)))))))))))))))))
s(s(half(x))) → s(p(s(s(p(p(half(s(s(p(p(s(s(p(s(p(x))))))))))))))))
0(sixtimes(x)) → s(p(s(p(s(s(s(s(s(0(s(p(s(p(x))))))))))))))
s(sixtimes(x)) → s(s(s(p(p(p(s(p(sixtimes(s(s(s(p(s(p(p(s(s(s(s(s(s(s(p(p(x)))))))))))))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → 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:
thrice(0(x1)) → p(s(p(p(p(s(s(s(0(p(s(p(s(x1)))))))))))))
thrice(s(x1)) → p(p(s(s(half(p(p(s(s(p(s(sixtimes(p(s(p(p(s(s(x1))))))))))))))))))
half(0(x1)) → p(p(s(s(p(s(0(p(s(s(s(s(x1))))))))))))
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(0(x1)) → p(s(p(s(0(s(s(s(s(s(p(s(p(s(x1))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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:
thrice(0(x1)) → p(s(p(p(p(s(s(s(0(p(s(p(s(x1)))))))))))))
thrice(s(x1)) → p(p(s(s(half(p(p(s(s(p(s(sixtimes(p(s(p(p(s(s(x1))))))))))))))))))
Used ordering:
Polynomial interpretation [25]:
POL(0(x1)) = x1
POL(half(x1)) = 2·x1
POL(p(x1)) = x1
POL(s(x1)) = x1
POL(sixtimes(x1)) = x1
POL(thrice(x1)) = 1 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
half(0(x1)) → p(p(s(s(p(s(0(p(s(s(s(s(x1))))))))))))
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(0(x1)) → p(s(p(s(0(s(s(s(s(s(p(s(p(s(x1))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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:
half(0(x1)) → p(p(s(s(p(s(0(p(s(s(s(s(x1))))))))))))
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(0(x1)) → p(s(p(s(0(s(s(s(s(s(p(s(p(s(x1))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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:
half(0(x1)) → p(p(s(s(p(s(0(p(s(s(s(s(x1))))))))))))
sixtimes(0(x1)) → p(s(p(s(0(s(s(s(s(s(p(s(p(s(x1))))))))))))))
0(x1) → x1
Used ordering:
Polynomial interpretation [25]:
POL(0(x1)) = 1 + x1
POL(half(x1)) = 2 + 2·x1
POL(p(x1)) = x1
POL(s(x1)) = x1
POL(sixtimes(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
SIXTIMES(s(x1)) → P(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))
HALF(s(s(x1))) → P(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
HALF(s(x1)) → P(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))
SIXTIMES(s(x1)) → P(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))
SIXTIMES(s(x1)) → P(p(s(s(s(x1)))))
HALF(s(x1)) → P(s(x1))
SIXTIMES(s(x1)) → P(s(s(s(x1))))
HALF(s(s(x1))) → P(p(s(s(half(p(p(s(s(p(s(x1)))))))))))
SIXTIMES(s(x1)) → P(p(p(s(s(s(x1))))))
HALF(s(s(x1))) → P(p(s(s(p(s(x1))))))
SIXTIMES(s(x1)) → P(s(p(p(p(s(s(s(x1))))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
SIXTIMES(s(x1)) → P(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1))))))))))))))))
HALF(s(x1)) → P(p(s(s(half(p(p(s(s(p(s(x1)))))))))))
P(p(s(x1))) → P(x1)
HALF(s(x1)) → P(s(s(p(s(x1)))))
HALF(s(x1)) → P(s(s(half(p(p(s(s(p(s(x1))))))))))
HALF(s(x1)) → P(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))
HALF(s(s(x1))) → P(s(s(p(s(x1)))))
HALF(s(s(x1))) → P(s(x1))
HALF(s(s(x1))) → P(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))
SIXTIMES(s(x1)) → SIXTIMES(p(s(p(p(p(s(s(s(x1)))))))))
HALF(s(x1)) → P(p(s(s(p(s(x1))))))
SIXTIMES(s(x1)) → P(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
HALF(s(x1)) → P(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
SIXTIMES(s(x1)) → P(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1))))))))))))))))))))))))
HALF(s(s(x1))) → P(s(s(half(p(p(s(s(p(s(x1))))))))))
The TRS R consists of the following rules:
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
SIXTIMES(s(x1)) → P(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))
HALF(s(s(x1))) → P(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
HALF(s(x1)) → P(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))
SIXTIMES(s(x1)) → P(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))
SIXTIMES(s(x1)) → P(p(s(s(s(x1)))))
HALF(s(x1)) → P(s(x1))
SIXTIMES(s(x1)) → P(s(s(s(x1))))
HALF(s(s(x1))) → P(p(s(s(half(p(p(s(s(p(s(x1)))))))))))
SIXTIMES(s(x1)) → P(p(p(s(s(s(x1))))))
HALF(s(s(x1))) → P(p(s(s(p(s(x1))))))
SIXTIMES(s(x1)) → P(s(p(p(p(s(s(s(x1))))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
SIXTIMES(s(x1)) → P(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1))))))))))))))))
HALF(s(x1)) → P(p(s(s(half(p(p(s(s(p(s(x1)))))))))))
P(p(s(x1))) → P(x1)
HALF(s(x1)) → P(s(s(p(s(x1)))))
HALF(s(x1)) → P(s(s(half(p(p(s(s(p(s(x1))))))))))
HALF(s(x1)) → P(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))
HALF(s(s(x1))) → P(s(s(p(s(x1)))))
HALF(s(s(x1))) → P(s(x1))
HALF(s(s(x1))) → P(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))
SIXTIMES(s(x1)) → SIXTIMES(p(s(p(p(p(s(s(s(x1)))))))))
HALF(s(x1)) → P(p(s(s(p(s(x1))))))
SIXTIMES(s(x1)) → P(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
HALF(s(x1)) → P(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
SIXTIMES(s(x1)) → P(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1))))))))))))))))))))))))
HALF(s(s(x1))) → P(s(s(half(p(p(s(s(p(s(x1))))))))))
The TRS R consists of the following rules:
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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 24 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:
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(p(p(p(s(s(s(x1)))))))))
The TRS R consists of the following rules:
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(p(p(p(s(s(s(x1)))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(p(p(p(s(s(s(x1)))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(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(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 SIXTIMES(s(x1)) → SIXTIMES(p(s(p(p(p(s(s(s(x1))))))))) at position [0] we obtained the following new rules:
SIXTIMES(s(x1)) → SIXTIMES(p(p(p(s(s(s(x1)))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(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(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 SIXTIMES(s(x1)) → SIXTIMES(p(p(p(s(s(s(x1))))))) at position [0,0,0] we obtained the following new rules:
SIXTIMES(s(x1)) → SIXTIMES(p(p(s(s(x1)))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(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(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 SIXTIMES(s(x1)) → SIXTIMES(p(p(s(s(x1))))) at position [0,0] we obtained the following new rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(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 SIXTIMES(s(x1)) → SIXTIMES(p(s(x1))) at position [0] we obtained the following new rules:
SIXTIMES(s(x1)) → SIXTIMES(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(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
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(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
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(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:
SIXTIMES(s(x1)) → SIXTIMES(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(SIXTIMES(x1)) = 2·x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ 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.
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
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(p(p(p(s(s(s(x1)))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(p(p(p(s(s(s(x1)))))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(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(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIXTIMES(s(x1)) → SIXTIMES(p(s(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(x1)))))
Q is empty.
We have to consider all (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
half(s(x1)) → p(s(p(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1)))))))))))))))))
half(s(s(x1))) → p(s(p(s(s(p(p(s(s(half(p(p(s(s(p(s(x1))))))))))))))))
sixtimes(s(x1)) → p(p(s(s(s(s(s(s(s(p(p(s(p(s(s(s(sixtimes(p(s(p(p(p(s(s(s(x1)))))))))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(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
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
Q is empty.
We have to consider all (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
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1)))))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(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 HALF(s(s(x1))) → HALF(p(p(s(s(p(s(x1))))))) at position [0,0] we obtained the following new rules:
HALF(s(s(x1))) → HALF(p(s(p(s(x1)))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(s(p(s(x1)))))
HALF(s(x1)) → HALF(p(p(s(s(p(s(x1)))))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(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 HALF(s(x1)) → HALF(p(p(s(s(p(s(x1))))))) at position [0,0] we obtained the following new rules:
HALF(s(x1)) → HALF(p(s(p(s(x1)))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(x1)) → HALF(p(s(p(s(x1)))))
HALF(s(s(x1))) → HALF(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
HALF(s(x1)) → HALF(p(s(p(s(x1)))))
HALF(s(s(x1))) → HALF(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 HALF(s(s(x1))) → HALF(p(s(p(s(x1))))) at position [0] we obtained the following new rules:
HALF(s(s(x1))) → HALF(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
HALF(s(x1)) → HALF(p(s(p(s(x1)))))
HALF(s(s(x1))) → HALF(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 HALF(s(x1)) → HALF(p(s(p(s(x1))))) at position [0] we obtained the following new rules:
HALF(s(x1)) → HALF(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(p(s(x1)))
HALF(s(x1)) → HALF(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 HALF(s(s(x1))) → HALF(p(s(x1))) at position [0] we obtained the following new rules:
HALF(s(s(x1))) → HALF(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(x1)
HALF(s(x1)) → HALF(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 HALF(s(x1)) → HALF(p(s(x1))) at position [0] we obtained the following new rules:
HALF(s(x1)) → HALF(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(x1)
HALF(s(x1)) → HALF(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(x1)
HALF(s(x1)) → HALF(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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x1))) → HALF(x1)
HALF(s(x1)) → HALF(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:
HALF(s(s(x1))) → HALF(x1)
HALF(s(x1)) → HALF(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(HALF(x1)) = 2·x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
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.