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:

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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MEM2(x, cons2(y, l)) -> EQ2(x, y)
*12(*2(x, y), z) -> *12(y, z)
LOG'1(11(x)) -> +12(log'1(x), 11(#))
PROD1(app2(l1, l2)) -> PROD1(l2)
IFINTER4(true, x, l1, l2) -> INTER2(l1, l2)
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
SUM1(app2(l1, l2)) -> +12(sum1(l1), sum1(l2))
INTER2(l1, cons2(x, l2)) -> MEM2(x, l1)
LOG'1(01(x)) -> LOG'1(x)
*12(x, +2(y, z)) -> *12(x, y)
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)
*12(*2(x, y), z) -> *12(x, *2(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)
INTER2(l1, app2(l2, l3)) -> INTER2(l1, l3)
+12(+2(x, y), z) -> +12(y, z)
PROD1(app2(l1, l2)) -> PROD1(l1)
*12(11(x), y) -> 011(*2(x, y))
*12(11(x), y) -> +12(01(*2(x, y)), y)
EQ2(01(x), 01(y)) -> EQ2(x, y)
APP2(cons2(x, l1), l2) -> APP2(l1, l2)
MEM2(x, cons2(y, l)) -> IF3(eq2(x, y), true, mem2(x, l))
LOG'1(01(x)) -> GE2(x, 11(#))
SUM1(app2(l1, l2)) -> SUM1(l1)
GE2(01(x), 11(y)) -> NOT1(ge2(y, x))
INTER2(app2(l1, l2), l3) -> INTER2(l1, l3)
LOG'1(01(x)) -> +12(log'1(x), 11(#))
PROD1(cons2(x, l)) -> *12(x, prod1(l))
INTER2(l1, app2(l2, l3)) -> INTER2(l1, l2)
PROD1(cons2(x, l)) -> PROD1(l)
*12(x, +2(y, z)) -> *12(x, z)
*12(01(x), y) -> *12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
IFINTER4(false, x, l1, l2) -> INTER2(l1, l2)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
SUM1(nil) -> 011(#)
EQ2(11(x), 11(y)) -> EQ2(x, y)
INTER2(app2(l1, l2), l3) -> INTER2(l2, l3)
-12(11(x), 11(y)) -> 011(-2(x, y))
EQ2(01(x), #) -> EQ2(x, #)
EQ2(#, 01(y)) -> EQ2(#, y)
+12(01(x), 01(y)) -> +12(x, y)
-12(01(x), 01(y)) -> 011(-2(x, y))
LOG1(x) -> LOG'1(x)
INTER2(l1, cons2(x, l2)) -> IFINTER4(mem2(x, l1), x, l2, l1)
*12(x, +2(y, z)) -> +12(*2(x, y), *2(x, z))
*12(11(x), y) -> *12(x, y)
+12(01(x), 01(y)) -> 011(+2(x, y))
MEM2(x, cons2(y, l)) -> MEM2(x, l)
LOG'1(11(x)) -> LOG'1(x)
SUM1(cons2(x, l)) -> +12(x, sum1(l))
LOG1(x) -> -12(log'1(x), 11(#))
GE2(11(x), 11(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> +12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
SUM1(app2(l1, l2)) -> SUM1(l2)
INTER2(l1, app2(l2, l3)) -> APP2(inter2(l1, l2), inter2(l1, l3))
+12(+2(x, y), z) -> +12(x, +2(y, z))
SUM1(cons2(x, l)) -> SUM1(l)
LOG'1(01(x)) -> IF3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
GE2(#, 01(x)) -> GE2(#, x)
INTER2(cons2(x, l1), l2) -> IFINTER4(mem2(x, l2), x, l1, l2)
PROD1(app2(l1, l2)) -> *12(prod1(l1), prod1(l2))
INTER2(app2(l1, l2), l3) -> APP2(inter2(l1, l3), inter2(l2, l3))
*12(01(x), y) -> 011(*2(x, y))
GE2(01(x), 01(y)) -> GE2(x, y)
INTER2(cons2(x, l1), l2) -> MEM2(x, l2)
-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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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:

MEM2(x, cons2(y, l)) -> EQ2(x, y)
*12(*2(x, y), z) -> *12(y, z)
LOG'1(11(x)) -> +12(log'1(x), 11(#))
PROD1(app2(l1, l2)) -> PROD1(l2)
IFINTER4(true, x, l1, l2) -> INTER2(l1, l2)
-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
SUM1(app2(l1, l2)) -> +12(sum1(l1), sum1(l2))
INTER2(l1, cons2(x, l2)) -> MEM2(x, l1)
LOG'1(01(x)) -> LOG'1(x)
*12(x, +2(y, z)) -> *12(x, y)
GE2(11(x), 01(y)) -> GE2(x, y)
GE2(01(x), 11(y)) -> GE2(y, x)
*12(*2(x, y), z) -> *12(x, *2(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)
INTER2(l1, app2(l2, l3)) -> INTER2(l1, l3)
+12(+2(x, y), z) -> +12(y, z)
PROD1(app2(l1, l2)) -> PROD1(l1)
*12(11(x), y) -> 011(*2(x, y))
*12(11(x), y) -> +12(01(*2(x, y)), y)
EQ2(01(x), 01(y)) -> EQ2(x, y)
APP2(cons2(x, l1), l2) -> APP2(l1, l2)
MEM2(x, cons2(y, l)) -> IF3(eq2(x, y), true, mem2(x, l))
LOG'1(01(x)) -> GE2(x, 11(#))
SUM1(app2(l1, l2)) -> SUM1(l1)
GE2(01(x), 11(y)) -> NOT1(ge2(y, x))
INTER2(app2(l1, l2), l3) -> INTER2(l1, l3)
LOG'1(01(x)) -> +12(log'1(x), 11(#))
PROD1(cons2(x, l)) -> *12(x, prod1(l))
INTER2(l1, app2(l2, l3)) -> INTER2(l1, l2)
PROD1(cons2(x, l)) -> PROD1(l)
*12(x, +2(y, z)) -> *12(x, z)
*12(01(x), y) -> *12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
IFINTER4(false, x, l1, l2) -> INTER2(l1, l2)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
SUM1(nil) -> 011(#)
EQ2(11(x), 11(y)) -> EQ2(x, y)
INTER2(app2(l1, l2), l3) -> INTER2(l2, l3)
-12(11(x), 11(y)) -> 011(-2(x, y))
EQ2(01(x), #) -> EQ2(x, #)
EQ2(#, 01(y)) -> EQ2(#, y)
+12(01(x), 01(y)) -> +12(x, y)
-12(01(x), 01(y)) -> 011(-2(x, y))
LOG1(x) -> LOG'1(x)
INTER2(l1, cons2(x, l2)) -> IFINTER4(mem2(x, l1), x, l2, l1)
*12(x, +2(y, z)) -> +12(*2(x, y), *2(x, z))
*12(11(x), y) -> *12(x, y)
+12(01(x), 01(y)) -> 011(+2(x, y))
MEM2(x, cons2(y, l)) -> MEM2(x, l)
LOG'1(11(x)) -> LOG'1(x)
SUM1(cons2(x, l)) -> +12(x, sum1(l))
LOG1(x) -> -12(log'1(x), 11(#))
GE2(11(x), 11(y)) -> GE2(x, y)
+12(11(x), 11(y)) -> +12(x, y)
-12(01(x), 01(y)) -> -12(x, y)
SUM1(app2(l1, l2)) -> SUM1(l2)
INTER2(l1, app2(l2, l3)) -> APP2(inter2(l1, l2), inter2(l1, l3))
+12(+2(x, y), z) -> +12(x, +2(y, z))
SUM1(cons2(x, l)) -> SUM1(l)
LOG'1(01(x)) -> IF3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
GE2(#, 01(x)) -> GE2(#, x)
INTER2(cons2(x, l1), l2) -> IFINTER4(mem2(x, l2), x, l1, l2)
PROD1(app2(l1, l2)) -> *12(prod1(l1), prod1(l2))
INTER2(app2(l1, l2), l3) -> APP2(inter2(l1, l3), inter2(l2, l3))
*12(01(x), y) -> 011(*2(x, y))
GE2(01(x), 01(y)) -> GE2(x, y)
INTER2(cons2(x, l1), l2) -> MEM2(x, l2)
-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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 14 SCCs with 26 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP2(cons2(x, l1), l2) -> APP2(l1, l2)

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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


APP2(cons2(x, l1), l2) -> APP2(l1, l2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1   
POL(cons2(x1, x2)) = 1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


GE2(#, 01(x)) -> GE2(#, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(01(x1)) = 1 + x1   
POL(GE2(x1, x2)) = x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   
POL(GE2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

EQ2(01(x), #) -> EQ2(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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


EQ2(01(x), #) -> EQ2(x, #)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(01(x1)) = 1 + x1   
POL(EQ2(x1, x2)) = x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

EQ2(#, 01(y)) -> EQ2(#, 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


EQ2(#, 01(y)) -> EQ2(#, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(01(x1)) = 1 + x1   
POL(EQ2(x1, x2)) = x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

EQ2(11(x), 11(y)) -> EQ2(x, y)
EQ2(01(x), 01(y)) -> EQ2(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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


EQ2(11(x), 11(y)) -> EQ2(x, y)
EQ2(01(x), 01(y)) -> EQ2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   
POL(EQ2(x1, x2)) = x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

MEM2(x, cons2(y, l)) -> MEM2(x, l)

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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MEM2(x, cons2(y, l)) -> MEM2(x, l)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MEM2(x1, x2)) = x2   
POL(cons2(x1, x2)) = 1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INTER2(l1, cons2(x, l2)) -> IFINTER4(mem2(x, l1), x, l2, l1)
INTER2(l1, app2(l2, l3)) -> INTER2(l1, l3)
INTER2(cons2(x, l1), l2) -> IFINTER4(mem2(x, l2), x, l1, l2)
INTER2(l1, app2(l2, l3)) -> INTER2(l1, l2)
IFINTER4(true, x, l1, l2) -> INTER2(l1, l2)
INTER2(app2(l1, l2), l3) -> INTER2(l2, l3)
INTER2(app2(l1, l2), l3) -> INTER2(l1, l3)
IFINTER4(false, x, l1, l2) -> INTER2(l1, l2)

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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


INTER2(l1, cons2(x, l2)) -> IFINTER4(mem2(x, l1), x, l2, l1)
INTER2(l1, app2(l2, l3)) -> INTER2(l1, l3)
INTER2(cons2(x, l1), l2) -> IFINTER4(mem2(x, l2), x, l1, l2)
INTER2(l1, app2(l2, l3)) -> INTER2(l1, l2)
INTER2(app2(l1, l2), l3) -> INTER2(l2, l3)
INTER2(app2(l1, l2), l3) -> INTER2(l1, l3)
The remaining pairs can at least be oriented weakly.

IFINTER4(true, x, l1, l2) -> INTER2(l1, l2)
IFINTER4(false, x, l1, l2) -> INTER2(l1, l2)
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(01(x1)) = 0   
POL(11(x1)) = 0   
POL(IFINTER4(x1, x2, x3, x4)) = x3 + x4   
POL(INTER2(x1, x2)) = x1 + x2   
POL(app2(x1, x2)) = 1 + x1 + x2   
POL(cons2(x1, x2)) = 1 + x2   
POL(eq2(x1, x2)) = 0   
POL(false) = 0   
POL(if3(x1, x2, x3)) = x2 + x3   
POL(mem2(x1, x2)) = 0   
POL(nil) = 0   
POL(true) = 0   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IFINTER4(true, x, l1, l2) -> INTER2(l1, l2)
IFINTER4(false, x, l1, l2) -> INTER2(l1, l2)

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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


-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)
The remaining pairs can at least be oriented weakly.

-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(-2(x1, x2)) = 0   
POL(-12(x1, x2)) = x2   
POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


-12(01(x), 11(y)) -> -12(-2(x, y), 11(#))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(-2(x1, x2)) = x1   
POL(-12(x1, x2)) = x1 + x2   
POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   

The following usable rules [14] were oriented:

-2(#, x) -> #
-2(01(x), 01(y)) -> 01(-2(x, y))
-2(11(x), 01(y)) -> 11(-2(x, y))
-2(x, #) -> x
01(#) -> #
-2(11(x), 11(y)) -> 01(-2(x, y))
-2(01(x), 11(y)) -> 11(-2(-2(x, y), 11(#)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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(+2(x, y), z) -> +12(y, z)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(x, +2(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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
The remaining pairs can at least be oriented weakly.

+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(+2(x1, x2)) = x1 + x2   
POL(+12(x1, x2)) = x1 + x2   
POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   

The following usable rules [14] were oriented:

+2(11(x), 01(y)) -> 11(+2(x, y))
+2(x, #) -> x
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(#, x) -> x
01(#) -> #
+2(+2(x, y), z) -> +2(x, +2(y, z))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(+2(x1, x2)) = 1 + x1 + x2   
POL(+12(x1, x2)) = x1   
POL(01(x1)) = 0   
POL(11(x1)) = 0   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SUM1(app2(l1, l2)) -> SUM1(l1)
SUM1(app2(l1, l2)) -> SUM1(l2)
SUM1(cons2(x, l)) -> SUM1(l)

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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


SUM1(app2(l1, l2)) -> SUM1(l1)
SUM1(app2(l1, l2)) -> SUM1(l2)
SUM1(cons2(x, l)) -> SUM1(l)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SUM1(x1)) = x1   
POL(app2(x1, x2)) = 1 + x1 + x2   
POL(cons2(x1, x2)) = 1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

*12(*2(x, y), z) -> *12(y, z)
*12(x, +2(y, z)) -> *12(x, z)
*12(11(x), y) -> *12(x, y)
*12(x, +2(y, z)) -> *12(x, y)
*12(01(x), y) -> *12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*12(*2(x, y), z) -> *12(y, z)
*12(11(x), y) -> *12(x, y)
*12(01(x), y) -> *12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, z))
The remaining pairs can at least be oriented weakly.

*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(*2(x1, x2)) = 1 + x1 + x2   
POL(*12(x1, x2)) = x1   
POL(+2(x1, x2)) = 0   
POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*12(x1, x2)) = x2   
POL(+2(x1, x2)) = 1 + x1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROD1(app2(l1, l2)) -> PROD1(l2)
PROD1(cons2(x, l)) -> PROD1(l)
PROD1(app2(l1, l2)) -> PROD1(l1)

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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROD1(app2(l1, l2)) -> PROD1(l2)
PROD1(cons2(x, l)) -> PROD1(l)
PROD1(app2(l1, l2)) -> PROD1(l1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROD1(x1)) = x1   
POL(app2(x1, x2)) = 1 + x1 + x2   
POL(cons2(x1, x2)) = 1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

LOG'1(11(x)) -> LOG'1(x)
LOG'1(01(x)) -> LOG'1(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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LOG'1(11(x)) -> LOG'1(x)
LOG'1(01(x)) -> LOG'1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(01(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   
POL(LOG'1(x1)) = x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(+2(x, y), z) -> +2(x, +2(y, z))
-2(#, x) -> #
-2(x, #) -> 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(true) -> false
not1(false) -> true
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(#, #) -> true
eq2(#, 11(y)) -> false
eq2(11(x), #) -> false
eq2(#, 01(y)) -> eq2(#, y)
eq2(01(x), #) -> eq2(x, #)
eq2(11(x), 11(y)) -> eq2(x, y)
eq2(01(x), 11(y)) -> false
eq2(11(x), 01(y)) -> false
eq2(01(x), 01(y)) -> eq2(x, 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(#, 01(x)) -> ge2(#, x)
ge2(#, 11(x)) -> false
log1(x) -> -2(log'1(x), 11(#))
log'1(#) -> #
log'1(11(x)) -> +2(log'1(x), 11(#))
log'1(01(x)) -> if3(ge2(x, 11(#)), +2(log'1(x), 11(#)), #)
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> if3(eq2(x, y), true, mem2(x, l))
inter2(x, nil) -> nil
inter2(nil, x) -> nil
inter2(app2(l1, l2), l3) -> app2(inter2(l1, l3), inter2(l2, l3))
inter2(l1, app2(l2, l3)) -> app2(inter2(l1, l2), inter2(l1, l3))
inter2(cons2(x, l1), l2) -> ifinter4(mem2(x, l2), x, l1, l2)
inter2(l1, cons2(x, l2)) -> ifinter4(mem2(x, l1), x, l2, l1)
ifinter4(true, x, l1, l2) -> cons2(x, inter2(l1, l2))
ifinter4(false, x, l1, l2) -> inter2(l1, l2)

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.