Termination w.r.t. Q of the given QTRS could not be shown:

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

(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(cons(x, l1), l2) → APP(l1, l2)
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, x2)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  cons(x2)
nil  =  nil

Lexicographic path order with status [LPO].
Quasi-Precedence:
[APP2, app2] > cons1
nil > cons1

Status:
APP2: [1,2]
cons1: [1]
app2: [1,2]
nil: []


The following usable rules [FROCOS05] were oriented:

app(app(l1, l2), l3) → app(l1, app(l2, l3))
app(cons(x, l1), l2) → cons(x, app(l1, l2))
app(nil, l) → l

(7) 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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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.

(11) 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: Combined order from the following AFS and order.
EQ(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(12) 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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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.

(16) 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(x1, x2)
IFMEM(x1, x2, x3)  =  IFMEM(x2, x3)
eq(x1, x2)  =  eq
false  =  false
s(x1)  =  s
0  =  0
true  =  true

Lexicographic path order with status [LPO].
Quasi-Precedence:
[MEM2, IFMEM2] > eq
cons2 > eq
[0, true] > false

Status:
eq: []
cons2: [1,2]
MEM2: [2,1]
true: []
false: []
s: []
IFMEM2: [2,1]
0: []


The following usable rules [FROCOS05] were oriented: none

(17) 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.

(18) DependencyGraphProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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.