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(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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented: none



↳ 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 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)) = x2   
POL(s1(x1)) = 1 + 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.


MEM2(x, cons2(y, l)) -> IFMEM3(eq2(x, y), x, l)
The remaining pairs can at least be oriented weakly.

IFMEM3(false, x, l) -> MEM2(x, l)
Used ordering: Polynomial interpretation [21]:

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

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
            ↳ 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(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) = 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 + x1 + x2   
POL(eq2(x1, x2)) = 0   
POL(false) = 0   
POL(ifmem3(x1, x2, x3)) = 0   
POL(mem2(x1, x2)) = 0   
POL(nil) = 0   
POL(s1(x1)) = 0   
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:

IFINTER4(true, x, l1, l2) -> INTER2(l1, l2)
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 0 SCCs with 2 less nodes.