Term Rewriting System R:
[x, y, l, l1, l2, l3]
if(true, x, y) -> x
if(false, x, y) -> y
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> mem(x, l)
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

EQ(s(x), s(y)) -> EQ(x, y)
APP(cons(x, l1), l2) -> APP(l1, l2)
APP(app(l1, l2), l3) -> APP(l1, app(l2, l3))
APP(app(l1, l2), l3) -> APP(l2, l3)
MEM(x, cons(y, l)) -> IFMEM(eq(x, y), x, l)
MEM(x, cons(y, l)) -> EQ(x, y)
IFMEM(false, x, l) -> MEM(x, l)
INTER(app(l1, l2), l3) -> APP(inter(l1, l3), inter(l2, l3))
INTER(app(l1, l2), l3) -> INTER(l1, l3)
INTER(app(l1, l2), l3) -> INTER(l2, l3)
INTER(l1, app(l2, l3)) -> APP(inter(l1, l2), inter(l1, l3))
INTER(l1, app(l2, l3)) -> INTER(l1, l2)
INTER(l1, app(l2, l3)) -> INTER(l1, l3)
INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
INTER(cons(x, l1), l2) -> MEM(x, l2)
INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
INTER(l1, cons(x, l2)) -> MEM(x, l1)
IFINTER(true, x, l1, l2) -> INTER(l1, l2)
IFINTER(false, x, l1, l2) -> INTER(l1, l2)

Furthermore, R contains four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

EQ(s(x), s(y)) -> EQ(x, y)

Rules:

if(true, x, y) -> x
if(false, x, y) -> y
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> mem(x, l)
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

The following dependency pair can be strictly oriented:

EQ(s(x), s(y)) -> EQ(x, y)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
EQ(x1, x2) -> EQ(x1, x2)
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

Rules:

if(true, x, y) -> x
if(false, x, y) -> y
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> mem(x, l)
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

APP(app(l1, l2), l3) -> APP(l2, l3)
APP(app(l1, l2), l3) -> APP(l1, app(l2, l3))
APP(cons(x, l1), l2) -> APP(l1, l2)

Rules:

if(true, x, y) -> x
if(false, x, y) -> y
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> mem(x, l)
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

The following dependency pairs can be strictly oriented:

APP(app(l1, l2), l3) -> APP(l2, l3)
APP(app(l1, l2), l3) -> APP(l1, app(l2, l3))
APP(cons(x, l1), l2) -> APP(l1, l2)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> x1
app(x1, x2) -> app(x1, x2)
cons(x1, x2) -> cons(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`           →DP Problem 6`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

Rules:

if(true, x, y) -> x
if(false, x, y) -> y
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> mem(x, l)
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pairs:

IFMEM(false, x, l) -> MEM(x, l)
MEM(x, cons(y, l)) -> IFMEM(eq(x, y), x, l)

Rules:

if(true, x, y) -> x
if(false, x, y) -> y
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> mem(x, l)
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

The following dependency pair can be strictly oriented:

MEM(x, cons(y, l)) -> IFMEM(eq(x, y), x, l)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
MEM(x1, x2) -> x2
cons(x1, x2) -> cons(x1, x2)
IFMEM(x1, x2, x3) -> x3

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`           →DP Problem 7`
`             ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

IFMEM(false, x, l) -> MEM(x, l)

Rules:

if(true, x, y) -> x
if(false, x, y) -> y
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> mem(x, l)
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

IFINTER(false, x, l1, l2) -> INTER(l1, l2)
INTER(l1, cons(x, l2)) -> IFINTER(mem(x, l1), x, l2, l1)
IFINTER(true, x, l1, l2) -> INTER(l1, l2)
INTER(cons(x, l1), l2) -> IFINTER(mem(x, l2), x, l1, l2)
INTER(l1, app(l2, l3)) -> INTER(l1, l3)
INTER(l1, app(l2, l3)) -> INTER(l1, l2)
INTER(app(l1, l2), l3) -> INTER(l2, l3)
INTER(app(l1, l2), l3) -> INTER(l1, l3)

Rules:

if(true, x, y) -> x
if(false, x, y) -> y
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
app(app(l1, l2), l3) -> app(l1, app(l2, l3))
mem(x, nil) -> false
mem(x, cons(y, l)) -> ifmem(eq(x, y), x, l)
ifmem(true, x, l) -> true
ifmem(false, x, l) -> mem(x, l)
inter(x, nil) -> nil
inter(nil, x) -> nil
inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3))
inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3))
inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2)
inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1)
ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2))
ifinter(false, x, l1, l2) -> inter(l1, l2)

Termination of R could not be shown.
Duration:
0:00 minutes