Termination w.r.t. Q proof of TRS/Cimeintersect.trs

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:

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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

APP₂(app₂(l1, l2), l3) -> APP₂(l2, l3)
APP₂(app₂(l1, l2), l3) -> APP₂(l1, app₂(l2, l3))

Used ordering: Polynomial interpretation [21]:

POL(APP₂(x₁, x₂)) = x₁
POL(app₂(x₁, x₂)) = x₁ + 2·x₂
POL(cons₂(x₁, x₂)) = 1 + x₁ + x₂
POL(nil) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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))

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 oriented strictly and are deleted.

APP₂(app₂(l1, l2), l3) -> APP₂(l2, l3)
APP₂(app₂(l1, l2), l3) -> APP₂(l1, app₂(l2, l3))

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

POL(APP₂(x₁, x₂)) = x₁
POL(app₂(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(cons₂(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(nil) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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 oriented strictly and are deleted.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

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

POL(EQ₂(x₁, x₂)) = 2·x₁ + x₂
POL(s₁(x₁)) = 2 + 2·x₁

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 oriented strictly and are deleted.

IFMEM₃(false, x, l) -> MEM₂(x, l)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(IFMEM₃(x₁, x₂, x₃)) = 2 + 2·x₁ + x₂ + 2·x₃
POL(MEM₂(x₁, x₂)) = 1 + x₁ + x₂
POL(cons₂(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(eq₂(x₁, x₂)) = x₂
POL(false) = 2
POL(s₁(x₁)) = 2 + 2·x₁
POL(true) = 1

The following usable rules [14] were oriented:

eq₂(0, s₁(x)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

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

            ↳ QDPOrderProof

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.
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(cons₂(x, l1), l2) -> IFINTER₄(mem₂(x, l2), x, l1, l2)

The remaining pairs can at least be oriented weakly.

INTER₂(l1, app₂(l2, l3)) -> INTER₂(l1, l3)
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)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(IFINTER₄(x₁, x₂, x₃, x₄)) = 1 + 2·x₂ + 2·x₃ + 2·x₄
POL(INTER₂(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(app₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(cons₂(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(eq₂(x₁, x₂)) = x₁ + x₂
POL(false) = 1
POL(ifmem₃(x₁, x₂, x₃)) = x₁ + 2·x₃
POL(mem₂(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(nil) = 2
POL(s₁(x₁)) = 2 + 2·x₁
POL(true) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

INTER₂(l1, app₂(l2, l3)) -> INTER₂(l1, l3)
INTER₂(l1, app₂(l2, l3)) -> INTER₂(l1, l2)
INTER₂(app₂(l1, l2), l3) -> INTER₂(l2, l3)
IFINTER₄(true, x, l1, l2) -> INTER₂(l1, l2)
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

INTER₂(l1, app₂(l2, l3)) -> INTER₂(l1, l3)
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 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 oriented strictly and are deleted.

INTER₂(l1, app₂(l2, l3)) -> INTER₂(l1, l3)
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.
none
Used ordering: Polynomial interpretation [21]:

POL(INTER₂(x₁, x₂)) = x₁ + 2·x₂
POL(app₂(x₁, x₂)) = 1 + 2·x₁ + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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.