Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/CimeSLASHintersect.trs

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = APP₂(x1, x2)
app₂(x1, x2) = app₂(x1, x2)
cons₂(x1, x2) = cons₁(x2)
nil = nil

Lexicographic Path Order [19].
Precedence:

[APP₂, app_2] > cons₁
nil > cons₁

The following usable rules [14] were oriented:

app₂(nil, l) -> l
app₂(cons₂(x, l1), l2) -> cons₂(x, app₂(l1, l2))
app₂(app₂(l1, l2), l3) -> app₂(l1, app₂(l2, l3))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
EQ₂(x1, x2) = EQ₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

MEM₂(x, cons₂(y, l)) -> IFMEM₃(eq₂(x, y), x, l)

The remaining pairs can at least by weakly be oriented.

IFMEM₃(false, x, l) -> MEM₂(x, l)

Used ordering: Combined order from the following AFS and order.
MEM₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
IFMEM₃(x1, x2, x3) = x3
eq₂(x1, x2) = eq₁(x2)
false = false
s₁(x1) = s
0 = 0
true = true

Lexicographic Path Order [19].
Precedence:

[eq₁, 0, true] > [false, s]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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 [13,14,18] 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.