Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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.
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(l1, l2), l3) -> APP2(l2, l3)
MEM2(x, cons2(y, l)) -> EQ2(x, y)
IFINTER4(true, x, l1, l2) -> INTER2(l1, l2)
INTER2(app2(l1, l2), l3) -> INTER2(l2, l3)
INTER2(app2(l1, l2), l3) -> INTER2(l1, l3)
INTER2(l1, cons2(x, l2)) -> MEM2(x, l1)
MEM2(x, cons2(y, l)) -> IFMEM3(eq2(x, y), x, l)
INTER2(l1, app2(l2, l3)) -> APP2(inter2(l1, l2), inter2(l1, l3))
IFMEM3(false, x, l) -> MEM2(x, l)
INTER2(l1, cons2(x, l2)) -> IFINTER4(mem2(x, l1), x, l2, l1)
EQ2(s1(x), s1(y)) -> EQ2(x, y)
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) -> APP2(inter2(l1, l3), inter2(l2, l3))
INTER2(cons2(x, l1), l2) -> MEM2(x, l2)
APP2(app2(l1, l2), l3) -> APP2(l1, app2(l2, l3))
IFINTER4(false, x, l1, l2) -> INTER2(l1, l2)
APP2(cons2(x, l1), l2) -> APP2(l1, l2)
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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:
APP2(app2(l1, l2), l3) -> APP2(l2, l3)
MEM2(x, cons2(y, l)) -> EQ2(x, y)
IFINTER4(true, x, l1, l2) -> INTER2(l1, l2)
INTER2(app2(l1, l2), l3) -> INTER2(l2, l3)
INTER2(app2(l1, l2), l3) -> INTER2(l1, l3)
INTER2(l1, cons2(x, l2)) -> MEM2(x, l1)
MEM2(x, cons2(y, l)) -> IFMEM3(eq2(x, y), x, l)
INTER2(l1, app2(l2, l3)) -> APP2(inter2(l1, l2), inter2(l1, l3))
IFMEM3(false, x, l) -> MEM2(x, l)
INTER2(l1, cons2(x, l2)) -> IFINTER4(mem2(x, l1), x, l2, l1)
EQ2(s1(x), s1(y)) -> EQ2(x, y)
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) -> APP2(inter2(l1, l3), inter2(l2, l3))
INTER2(cons2(x, l1), l2) -> MEM2(x, l2)
APP2(app2(l1, l2), l3) -> APP2(l1, app2(l2, l3))
IFINTER4(false, x, l1, l2) -> INTER2(l1, l2)
APP2(cons2(x, l1), l2) -> APP2(l1, l2)
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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 contains 4 SCCs with 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(l1, l2), l3) -> APP2(l2, l3)
APP2(app2(l1, l2), l3) -> APP2(l1, app2(l2, l3))
APP2(cons2(x, l1), l2) -> APP2(l1, l2)
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(app2(l1, l2), l3) -> APP2(l2, l3)
APP2(app2(l1, l2), l3) -> APP2(l1, app2(l2, l3))
Used argument filtering: APP2(x1, x2) = x1
app2(x1, x2) = app2(x1, x2)
cons2(x1, x2) = x2
nil = nil
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ 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:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(cons2(x, l1), l2) -> APP2(l1, l2)
Used argument filtering: APP2(x1, x2) = x1
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
EQ2(s1(x), s1(y)) -> EQ2(x, y)
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
EQ2(s1(x), s1(y)) -> EQ2(x, y)
Used argument filtering: EQ2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MEM2(x, cons2(y, l)) -> IFMEM3(eq2(x, y), x, l)
IFMEM3(false, x, l) -> MEM2(x, l)
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
MEM2(x, cons2(y, l)) -> IFMEM3(eq2(x, y), x, l)
Used argument filtering: MEM2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
IFMEM3(x1, x2, x3) = x3
eq2(x1, x2) = eq
0 = 0
true = true
s1(x1) = s
false = false
Used ordering: Quasi Precedence:
[eq, false] > true
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IFMEM3(false, x, l) -> MEM2(x, l)
The TRS R consists of the following rules:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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 contains 0 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ 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:
if3(true, x, y) -> x
if3(false, x, y) -> y
eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
app2(app2(l1, l2), l3) -> app2(l1, app2(l2, l3))
mem2(x, nil) -> false
mem2(x, cons2(y, l)) -> ifmem3(eq2(x, y), x, l)
ifmem3(true, x, l) -> true
ifmem3(false, x, l) -> 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.