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:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
BS1(n3(x, y, z)) -> GE2(x, max1(y))
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)
SIZE1(n3(x, y, z)) -> +12(+2(size1(x), size1(y)), 11(#))
+12(x, +2(y, z)) -> +12(+2(x, y), z)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(x, y)
SIZE1(n3(x, y, z)) -> +12(size1(x), size1(y))
BS1(n3(x, y, z)) -> BS1(z)
WB1(n3(x, y, z)) -> GE2(11(#), -2(size1(y), size1(z)))
WB1(n3(x, y, z)) -> GE2(11(#), -2(size1(z), size1(y)))
WB1(n3(x, y, z)) -> AND2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
BS1(n3(x, y, z)) -> AND2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
GE2(01(x), 11(y)) -> NOT1(ge2(y, x))
SIZE1(n3(x, y, z)) -> SIZE1(y)
+12(x, +2(y, z)) -> +12(x, y)
SIZE1(n3(x, y, z)) -> SIZE1(x)
WB1(n3(x, y, z)) -> SIZE1(y)
WB1(n3(x, y, z)) -> SIZE1(z)
BS1(n3(x, y, z)) -> BS1(y)
MIN1(n3(x, y, z)) -> MIN1(y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
WB1(n3(x, y, z)) -> AND2(wb1(y), wb1(z))
-12(11(x), 11(y)) -> 011(-2(x, y))
BS1(n3(x, y, z)) -> MAX1(y)
WB1(n3(x, y, z)) -> GE2(size1(y), size1(z))
+12(01(x), 01(y)) -> +12(x, y)
-12(01(x), 01(y)) -> 011(-2(x, y))
MAX1(n3(x, y, z)) -> MAX1(z)
WB1(n3(x, y, z)) -> -12(size1(y), size1(z))
WB1(n3(x, y, z)) -> -12(size1(z), size1(y))
BS1(n3(x, y, z)) -> AND2(ge2(x, max1(y)), ge2(min1(z), x))
+12(01(x), 01(y)) -> 011(+2(x, y))
WB1(n3(x, y, z)) -> WB1(z)
GE2(11(x), 11(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> +12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
WB1(n3(x, y, z)) -> WB1(y)
GE2(#, 01(x)) -> GE2(#, x)
BS1(n3(x, y, z)) -> MIN1(z)
BS1(n3(x, y, z)) -> GE2(min1(z), x)
BS1(n3(x, y, z)) -> AND2(bs1(y), bs1(z))
WB1(n3(x, y, z)) -> IF3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y))))
GE2(01(x), 01(y)) -> GE2(x, y)
-12(11(x), 11(y)) -> -12(x, y)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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:
BS1(n3(x, y, z)) -> GE2(x, max1(y))
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)
SIZE1(n3(x, y, z)) -> +12(+2(size1(x), size1(y)), 11(#))
+12(x, +2(y, z)) -> +12(+2(x, y), z)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(x, y)
SIZE1(n3(x, y, z)) -> +12(size1(x), size1(y))
BS1(n3(x, y, z)) -> BS1(z)
WB1(n3(x, y, z)) -> GE2(11(#), -2(size1(y), size1(z)))
WB1(n3(x, y, z)) -> GE2(11(#), -2(size1(z), size1(y)))
WB1(n3(x, y, z)) -> AND2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
BS1(n3(x, y, z)) -> AND2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
GE2(01(x), 11(y)) -> NOT1(ge2(y, x))
SIZE1(n3(x, y, z)) -> SIZE1(y)
+12(x, +2(y, z)) -> +12(x, y)
SIZE1(n3(x, y, z)) -> SIZE1(x)
WB1(n3(x, y, z)) -> SIZE1(y)
WB1(n3(x, y, z)) -> SIZE1(z)
BS1(n3(x, y, z)) -> BS1(y)
MIN1(n3(x, y, z)) -> MIN1(y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
WB1(n3(x, y, z)) -> AND2(wb1(y), wb1(z))
-12(11(x), 11(y)) -> 011(-2(x, y))
BS1(n3(x, y, z)) -> MAX1(y)
WB1(n3(x, y, z)) -> GE2(size1(y), size1(z))
+12(01(x), 01(y)) -> +12(x, y)
-12(01(x), 01(y)) -> 011(-2(x, y))
MAX1(n3(x, y, z)) -> MAX1(z)
WB1(n3(x, y, z)) -> -12(size1(y), size1(z))
WB1(n3(x, y, z)) -> -12(size1(z), size1(y))
BS1(n3(x, y, z)) -> AND2(ge2(x, max1(y)), ge2(min1(z), x))
+12(01(x), 01(y)) -> 011(+2(x, y))
WB1(n3(x, y, z)) -> WB1(z)
GE2(11(x), 11(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> +12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
WB1(n3(x, y, z)) -> WB1(y)
GE2(#, 01(x)) -> GE2(#, x)
BS1(n3(x, y, z)) -> MIN1(z)
BS1(n3(x, y, z)) -> GE2(min1(z), x)
BS1(n3(x, y, z)) -> AND2(bs1(y), bs1(z))
WB1(n3(x, y, z)) -> IF3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y))))
GE2(01(x), 01(y)) -> GE2(x, y)
-12(11(x), 11(y)) -> -12(x, y)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 9 SCCs with 24 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MAX1(n3(x, y, z)) -> MAX1(z)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.
MAX1(n3(x, y, z)) -> MAX1(z)
Used argument filtering: MAX1(x1) = x1
n3(x1, x2, x3) = n1(x3)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MIN1(n3(x, y, z)) -> MIN1(y)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.
MIN1(n3(x, y, z)) -> MIN1(y)
Used argument filtering: MIN1(x1) = x1
n3(x1, x2, x3) = n1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GE2(#, 01(x)) -> GE2(#, x)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.
GE2(#, 01(x)) -> GE2(#, x)
Used argument filtering: GE2(x1, x2) = x2
01(x1) = 01(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GE2(11(x), 11(y)) -> GE2(x, y)
GE2(01(x), 01(y)) -> GE2(x, y)
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
BS1(n3(x, y, z)) -> BS1(y)
BS1(n3(x, y, z)) -> BS1(z)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.
BS1(n3(x, y, z)) -> BS1(y)
BS1(n3(x, y, z)) -> BS1(z)
Used argument filtering: BS1(x1) = x1
n3(x1, x2, x3) = n2(x2, x3)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
-12(11(x), 11(y)) -> -12(x, y)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.
-12(11(x), 01(y)) -> -12(x, y)
-12(01(x), 11(y)) -> -12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
-12(11(x), 11(y)) -> -12(x, y)
Used argument filtering: -12(x1, x2) = x2
01(x1) = 01(x1)
11(x1) = 11(x1)
# = #
-2(x1, x2) = x1
Used ordering: Quasi Precedence:
[0_1, 1_1] > #
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
Used argument filtering: -12(x1, x2) = x1
01(x1) = 01(x1)
-2(x1, x2) = x1
# = #
11(x1) = 11(x1)
Used ordering: Quasi Precedence:
[0_1, 1_1]
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SIZE1(n3(x, y, z)) -> SIZE1(y)
SIZE1(n3(x, y, z)) -> SIZE1(x)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.
SIZE1(n3(x, y, z)) -> SIZE1(y)
SIZE1(n3(x, y, z)) -> SIZE1(x)
Used argument filtering: SIZE1(x1) = x1
n3(x1, x2, x3) = n2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
WB1(n3(x, y, z)) -> WB1(y)
WB1(n3(x, y, z)) -> WB1(z)
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.
WB1(n3(x, y, z)) -> WB1(y)
WB1(n3(x, y, z)) -> WB1(z)
Used argument filtering: WB1(x1) = x1
n3(x1, x2, x3) = n2(x2, x3)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
-2(x, #) -> x
-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(11(x), 11(y)) -> 01(-2(x, y))
not1(false) -> true
not1(true) -> false
and2(x, true) -> x
and2(x, false) -> false
if3(true, x, y) -> x
if3(false, x, y) -> y
ge2(01(x), 01(y)) -> ge2(x, y)
ge2(01(x), 11(y)) -> not1(ge2(y, x))
ge2(11(x), 01(y)) -> ge2(x, y)
ge2(11(x), 11(y)) -> ge2(x, y)
ge2(x, #) -> true
ge2(#, 11(x)) -> false
ge2(#, 01(x)) -> ge2(#, x)
val1(l1(x)) -> x
val1(n3(x, y, z)) -> x
min1(l1(x)) -> x
min1(n3(x, y, z)) -> min1(y)
max1(l1(x)) -> x
max1(n3(x, y, z)) -> max1(z)
bs1(l1(x)) -> true
bs1(n3(x, y, z)) -> and2(and2(ge2(x, max1(y)), ge2(min1(z), x)), and2(bs1(y), bs1(z)))
size1(l1(x)) -> 11(#)
size1(n3(x, y, z)) -> +2(+2(size1(x), size1(y)), 11(#))
wb1(l1(x)) -> true
wb1(n3(x, y, z)) -> and2(if3(ge2(size1(y), size1(z)), ge2(11(#), -2(size1(y), size1(z))), ge2(11(#), -2(size1(z), size1(y)))), and2(wb1(y), wb1(z)))
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.