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

(0) Obligation:

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

and(true, y) → y
and(false, y) → false
eq(nil, nil) → true
eq(cons(t, l), nil) → false
eq(nil, cons(t, l)) → false
eq(cons(t, l), cons(t', l')) → and(eq(t, t'), eq(l, l'))
eq(var(l), var(l')) → eq(l, l')
eq(var(l), apply(t, s)) → false
eq(var(l), lambda(x, t)) → false
eq(apply(t, s), var(l)) → false
eq(apply(t, s), apply(t', s')) → and(eq(t, t'), eq(s, s'))
eq(apply(t, s), lambda(x, t)) → false
eq(lambda(x, t), var(l)) → false
eq(lambda(x, t), apply(t, s)) → false
eq(lambda(x, t), lambda(x', t')) → and(eq(x, x'), eq(t, t'))
if(true, var(k), var(l')) → var(k)
if(false, var(k), var(l')) → var(l')
ren(var(l), var(k), var(l')) → if(eq(l, l'), var(k), var(l'))
ren(x, y, apply(t, s)) → apply(ren(x, y, t), ren(x, y, s))
ren(x, y, lambda(z, t)) → lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))

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(cons(t, l), cons(t', l')) → AND(eq(t, t'), eq(l, l'))
EQ(cons(t, l), cons(t', l')) → EQ(t, t')
EQ(cons(t, l), cons(t', l')) → EQ(l, l')
EQ(var(l), var(l')) → EQ(l, l')
EQ(apply(t, s), apply(t', s')) → AND(eq(t, t'), eq(s, s'))
EQ(apply(t, s), apply(t', s')) → EQ(t, t')
EQ(apply(t, s), apply(t', s')) → EQ(s, s')
EQ(lambda(x, t), lambda(x', t')) → AND(eq(x, x'), eq(t, t'))
EQ(lambda(x, t), lambda(x', t')) → EQ(x, x')
EQ(lambda(x, t), lambda(x', t')) → EQ(t, t')
REN(var(l), var(k), var(l')) → IF(eq(l, l'), var(k), var(l'))
REN(var(l), var(k), var(l')) → EQ(l, l')
REN(x, y, apply(t, s)) → REN(x, y, t)
REN(x, y, apply(t, s)) → REN(x, y, s)
REN(x, y, lambda(z, t)) → REN(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t))
REN(x, y, lambda(z, t)) → REN(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)

The TRS R consists of the following rules:

and(true, y) → y
and(false, y) → false
eq(nil, nil) → true
eq(cons(t, l), nil) → false
eq(nil, cons(t, l)) → false
eq(cons(t, l), cons(t', l')) → and(eq(t, t'), eq(l, l'))
eq(var(l), var(l')) → eq(l, l')
eq(var(l), apply(t, s)) → false
eq(var(l), lambda(x, t)) → false
eq(apply(t, s), var(l)) → false
eq(apply(t, s), apply(t', s')) → and(eq(t, t'), eq(s, s'))
eq(apply(t, s), lambda(x, t)) → false
eq(lambda(x, t), var(l)) → false
eq(lambda(x, t), apply(t, s)) → false
eq(lambda(x, t), lambda(x', t')) → and(eq(x, x'), eq(t, t'))
if(true, var(k), var(l')) → var(k)
if(false, var(k), var(l')) → var(l')
ren(var(l), var(k), var(l')) → if(eq(l, l'), var(k), var(l'))
ren(x, y, apply(t, s)) → apply(ren(x, y, t), ren(x, y, s))
ren(x, y, lambda(z, t)) → lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))

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 2 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

EQ(cons(t, l), cons(t', l')) → EQ(l, l')
EQ(cons(t, l), cons(t', l')) → EQ(t, t')
EQ(var(l), var(l')) → EQ(l, l')
EQ(apply(t, s), apply(t', s')) → EQ(t, t')
EQ(apply(t, s), apply(t', s')) → EQ(s, s')
EQ(lambda(x, t), lambda(x', t')) → EQ(x, x')
EQ(lambda(x, t), lambda(x', t')) → EQ(t, t')

The TRS R consists of the following rules:

and(true, y) → y
and(false, y) → false
eq(nil, nil) → true
eq(cons(t, l), nil) → false
eq(nil, cons(t, l)) → false
eq(cons(t, l), cons(t', l')) → and(eq(t, t'), eq(l, l'))
eq(var(l), var(l')) → eq(l, l')
eq(var(l), apply(t, s)) → false
eq(var(l), lambda(x, t)) → false
eq(apply(t, s), var(l)) → false
eq(apply(t, s), apply(t', s')) → and(eq(t, t'), eq(s, s'))
eq(apply(t, s), lambda(x, t)) → false
eq(lambda(x, t), var(l)) → false
eq(lambda(x, t), apply(t, s)) → false
eq(lambda(x, t), lambda(x', t')) → and(eq(x, x'), eq(t, t'))
if(true, var(k), var(l')) → var(k)
if(false, var(k), var(l')) → var(l')
ren(var(l), var(k), var(l')) → if(eq(l, l'), var(k), var(l'))
ren(x, y, apply(t, s)) → apply(ren(x, y, t), ren(x, y, s))
ren(x, y, lambda(z, t)) → lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))

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.


EQ(cons(t, l), cons(t', l')) → EQ(l, l')
EQ(cons(t, l), cons(t', l')) → EQ(t, t')
EQ(apply(t, s), apply(t', s')) → EQ(t, t')
EQ(apply(t, s), apply(t', s')) → EQ(s, s')
EQ(lambda(x, t), lambda(x', t')) → EQ(x, x')
EQ(lambda(x, t), lambda(x', t')) → EQ(t, t')
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
EQ(x1, x2)  =  EQ(x2)

Tags:
EQ has tags [1,0]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
cons(x1, x2)  =  cons(x1, x2)
var(x1)  =  x1
apply(x1, x2)  =  apply(x1, x2)
lambda(x1, x2)  =  lambda(x1, x2)

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

EQ(var(l), var(l')) → EQ(l, l')

The TRS R consists of the following rules:

and(true, y) → y
and(false, y) → false
eq(nil, nil) → true
eq(cons(t, l), nil) → false
eq(nil, cons(t, l)) → false
eq(cons(t, l), cons(t', l')) → and(eq(t, t'), eq(l, l'))
eq(var(l), var(l')) → eq(l, l')
eq(var(l), apply(t, s)) → false
eq(var(l), lambda(x, t)) → false
eq(apply(t, s), var(l)) → false
eq(apply(t, s), apply(t', s')) → and(eq(t, t'), eq(s, s'))
eq(apply(t, s), lambda(x, t)) → false
eq(lambda(x, t), var(l)) → false
eq(lambda(x, t), apply(t, s)) → false
eq(lambda(x, t), lambda(x', t')) → and(eq(x, x'), eq(t, t'))
if(true, var(k), var(l')) → var(k)
if(false, var(k), var(l')) → var(l')
ren(var(l), var(k), var(l')) → if(eq(l, l'), var(k), var(l'))
ren(x, y, apply(t, s)) → apply(ren(x, y, t), ren(x, y, s))
ren(x, y, lambda(z, t)) → lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))

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.


EQ(var(l), var(l')) → EQ(l, l')
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
EQ(x1, x2)  =  EQ(x2)

Tags:
EQ has tags [1,1]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

and(true, y) → y
and(false, y) → false
eq(nil, nil) → true
eq(cons(t, l), nil) → false
eq(nil, cons(t, l)) → false
eq(cons(t, l), cons(t', l')) → and(eq(t, t'), eq(l, l'))
eq(var(l), var(l')) → eq(l, l')
eq(var(l), apply(t, s)) → false
eq(var(l), lambda(x, t)) → false
eq(apply(t, s), var(l)) → false
eq(apply(t, s), apply(t', s')) → and(eq(t, t'), eq(s, s'))
eq(apply(t, s), lambda(x, t)) → false
eq(lambda(x, t), var(l)) → false
eq(lambda(x, t), apply(t, s)) → false
eq(lambda(x, t), lambda(x', t')) → and(eq(x, x'), eq(t, t'))
if(true, var(k), var(l')) → var(k)
if(false, var(k), var(l')) → var(l')
ren(var(l), var(k), var(l')) → if(eq(l, l'), var(k), var(l'))
ren(x, y, apply(t, s)) → apply(ren(x, y, t), ren(x, y, s))
ren(x, y, lambda(z, t)) → lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))

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:

REN(x, y, apply(t, s)) → REN(x, y, s)
REN(x, y, apply(t, s)) → REN(x, y, t)
REN(x, y, lambda(z, t)) → REN(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t))
REN(x, y, lambda(z, t)) → REN(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)

The TRS R consists of the following rules:

and(true, y) → y
and(false, y) → false
eq(nil, nil) → true
eq(cons(t, l), nil) → false
eq(nil, cons(t, l)) → false
eq(cons(t, l), cons(t', l')) → and(eq(t, t'), eq(l, l'))
eq(var(l), var(l')) → eq(l, l')
eq(var(l), apply(t, s)) → false
eq(var(l), lambda(x, t)) → false
eq(apply(t, s), var(l)) → false
eq(apply(t, s), apply(t', s')) → and(eq(t, t'), eq(s, s'))
eq(apply(t, s), lambda(x, t)) → false
eq(lambda(x, t), var(l)) → false
eq(lambda(x, t), apply(t, s)) → false
eq(lambda(x, t), lambda(x', t')) → and(eq(x, x'), eq(t, t'))
if(true, var(k), var(l')) → var(k)
if(false, var(k), var(l')) → var(l')
ren(var(l), var(k), var(l')) → if(eq(l, l'), var(k), var(l'))
ren(x, y, apply(t, s)) → apply(ren(x, y, t), ren(x, y, s))
ren(x, y, lambda(z, t)) → lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))

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.


REN(x, y, apply(t, s)) → REN(x, y, s)
REN(x, y, apply(t, s)) → REN(x, y, t)
REN(x, y, lambda(z, t)) → REN(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t))
REN(x, y, lambda(z, t)) → REN(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
REN(x1, x2, x3)  =  REN(x3)

Tags:
REN has tags [0,1,1]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
apply(x1, x2)  =  apply(x1, x2)
lambda(x1, x2)  =  lambda(x2)
ren(x1, x2, x3)  =  x3
var(x1)  =  var
cons(x1, x2)  =  cons(x1)
nil  =  nil
if(x1, x2, x3)  =  x3
eq(x1, x2)  =  eq(x1, x2)
true  =  true
false  =  false
and(x1, x2)  =  x2

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:

ren(var(l), var(k), var(l')) → if(eq(l, l'), var(k), var(l'))
ren(x, y, apply(t, s)) → apply(ren(x, y, t), ren(x, y, s))
ren(x, y, lambda(z, t)) → lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))
if(true, var(k), var(l')) → var(k)
if(false, var(k), var(l')) → var(l')

(14) Obligation:

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

and(true, y) → y
and(false, y) → false
eq(nil, nil) → true
eq(cons(t, l), nil) → false
eq(nil, cons(t, l)) → false
eq(cons(t, l), cons(t', l')) → and(eq(t, t'), eq(l, l'))
eq(var(l), var(l')) → eq(l, l')
eq(var(l), apply(t, s)) → false
eq(var(l), lambda(x, t)) → false
eq(apply(t, s), var(l)) → false
eq(apply(t, s), apply(t', s')) → and(eq(t, t'), eq(s, s'))
eq(apply(t, s), lambda(x, t)) → false
eq(lambda(x, t), var(l)) → false
eq(lambda(x, t), apply(t, s)) → false
eq(lambda(x, t), lambda(x', t')) → and(eq(x, x'), eq(t, t'))
if(true, var(k), var(l')) → var(k)
if(false, var(k), var(l')) → var(l')
ren(var(l), var(k), var(l')) → if(eq(l, l'), var(k), var(l'))
ren(x, y, apply(t, s)) → apply(ren(x, y, t), ren(x, y, s))
ren(x, y, lambda(z, t)) → lambda(var(cons(x, cons(y, cons(lambda(z, t), nil)))), ren(x, y, ren(z, var(cons(x, cons(y, cons(lambda(z, t), nil)))), t)))

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