Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔)
        ↳11 TRUE
    ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 TRUE
    ↳17 QDP
        ↳18 QDPOrderProof (⇔)
        ↳19 QDP
        ↳20 DependencyGraphProof (⇔)
        ↳21 TRUE
    ↳22 QDP
        ↳23 QDPOrderProof (⇔)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔)
        ↳26 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following 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)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(l1, l2), l3) → APP(l1, app(l2, l3))
APP(cons(x, l1), l2) → APP(l1, l2)
APP(app(l1, l2), l3) → APP(l2, l3)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(l1, l2), l3) → APP(l1, app(l2, l3))
APP(app(l1, l2), l3) → APP(l2, l3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  x2
nil  =  nil

Recursive path order with status [RPO].
Quasi-Precedence:
[APP1, app2]

Status:
APP1: multiset
app2: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

APP(cons(x, l1), l2) → APP(l1, l2)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(cons(x, l1), l2) → APP(l1, l2)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:
cons2 > APP2

Status:
APP2: [2,1]
cons2: multiset


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQ(s(x), s(y)) → EQ(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:
[EQ2, s1]

Status:
EQ2: [1,2]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

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

MEM(x, cons(y, l)) → IFMEM(eq(x, y), x, l)
IFMEM(false, x, l) → MEM(x, l)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MEM(x, cons(y, l)) → IFMEM(eq(x, y), x, l)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MEM(x1, x2)  =  MEM(x1, x2)
cons(x1, x2)  =  cons(x2)
IFMEM(x1, x2, x3)  =  IFMEM(x2, x3)
eq(x1, x2)  =  eq(x2)
false  =  false
0  =  0
true  =  true
s(x1)  =  s

Recursive path order with status [RPO].
Quasi-Precedence:
[false, 0, true] > [MEM2, IFMEM2]

Status:
MEM2: [2,1]
cons1: multiset
IFMEM2: [2,1]
eq1: multiset
false: multiset
0: multiset
true: multiset
s: multiset


The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

IFMEM(false, x, l) → MEM(x, l)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(20) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(21) TRUE

(22) Obligation:

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

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

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


INTER(app(l1, l2), l3) → INTER(l2, l3)
INTER(app(l1, l2), l3) → 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)
IFINTER(true, x, l1, l2) → INTER(l1, l2)
INTER(l1, cons(x, l2)) → IFINTER(mem(x, l1), x, l2, l1)
IFINTER(false, x, l1, l2) → INTER(l1, l2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INTER(x1, x2)  =  INTER(x1, x2)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
IFINTER(x1, x2, x3, x4)  =  IFINTER(x1, x2, x3, x4)
mem(x1, x2)  =  mem(x1)
true  =  true
false  =  false
nil  =  nil
ifmem(x1, x2, x3)  =  ifmem(x2)
eq(x1, x2)  =  eq(x1, x2)
0  =  0
s(x1)  =  s

Recursive path order with status [RPO].
Quasi-Precedence:
[cons2, mem1, ifmem1] > [INTER2, IFINTER4, false, nil]
[cons2, mem1, ifmem1] > true
[cons2, mem1, ifmem1] > eq2
s > [INTER2, IFINTER4, false, nil]
s > eq2

Status:
INTER2: multiset
app2: [1,2]
cons2: [1,2]
IFINTER4: multiset
mem1: [1]
true: multiset
false: multiset
nil: multiset
ifmem1: [1]
eq2: [2,1]
0: multiset
s: []


The following usable rules [FROCOS05] were oriented:

mem(x, nil) → false
mem(x, cons(y, l)) → ifmem(eq(x, y), x, l)
ifmem(false, x, l) → mem(x, l)
ifmem(true, x, l) → true

(24) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE