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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 PisEmptyProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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)))

The set Q consists of the following terms:

and(true, x0)
and(false, x0)
eq(nil, nil)
eq(cons(x0, x1), nil)
eq(nil, cons(x0, x1))
eq(cons(x0, x1), cons(x2, x3))
eq(var(x0), var(x1))
eq(var(x0), apply(x1, x2))
eq(var(x0), lambda(x1, x2))
eq(apply(x0, x1), var(x2))
eq(apply(x0, x1), apply(x2, x3))
eq(apply(x0, x1), lambda(x2, x0))
eq(lambda(x0, x1), var(x2))
eq(lambda(x0, x1), apply(x1, x2))
eq(lambda(x0, x1), lambda(x2, x3))
if(true, var(x0), var(x1))
if(false, var(x0), var(x1))
ren(var(x0), var(x1), var(x2))
ren(x0, x1, apply(x2, x3))
ren(x0, x1, lambda(x2, x3))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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)))

The set Q consists of the following terms:

and(true, x0)
and(false, x0)
eq(nil, nil)
eq(cons(x0, x1), nil)
eq(nil, cons(x0, x1))
eq(cons(x0, x1), cons(x2, x3))
eq(var(x0), var(x1))
eq(var(x0), apply(x1, x2))
eq(var(x0), lambda(x1, x2))
eq(apply(x0, x1), var(x2))
eq(apply(x0, x1), apply(x2, x3))
eq(apply(x0, x1), lambda(x2, x0))
eq(lambda(x0, x1), var(x2))
eq(lambda(x0, x1), apply(x1, x2))
eq(lambda(x0, x1), lambda(x2, x3))
if(true, var(x0), var(x1))
if(false, var(x0), var(x1))
ren(var(x0), var(x1), var(x2))
ren(x0, x1, apply(x2, x3))
ren(x0, x1, lambda(x2, x3))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) 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)))

The set Q consists of the following terms:

and(true, x0)
and(false, x0)
eq(nil, nil)
eq(cons(x0, x1), nil)
eq(nil, cons(x0, x1))
eq(cons(x0, x1), cons(x2, x3))
eq(var(x0), var(x1))
eq(var(x0), apply(x1, x2))
eq(var(x0), lambda(x1, x2))
eq(apply(x0, x1), var(x2))
eq(apply(x0, x1), apply(x2, x3))
eq(apply(x0, x1), lambda(x2, x0))
eq(lambda(x0, x1), var(x2))
eq(lambda(x0, x1), apply(x1, x2))
eq(lambda(x0, x1), lambda(x2, x3))
if(true, var(x0), var(x1))
if(false, var(x0), var(x1))
ren(var(x0), var(x1), var(x2))
ren(x0, x1, apply(x2, x3))
ren(x0, x1, lambda(x2, x3))

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(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: Combined order from the following AFS and order.
EQ(x1, x2)  =  EQ(x2)
cons(x1, x2)  =  cons(x1, x2)
var(x1)  =  x1
apply(x1, x2)  =  apply(x1, x2)
lambda(x1, x2)  =  lambda(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
cons2 > EQ1
apply2 > EQ1
lambda2 > EQ1

The following usable rules [FROCOS05] were oriented: none

(9) 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)))

The set Q consists of the following terms:

and(true, x0)
and(false, x0)
eq(nil, nil)
eq(cons(x0, x1), nil)
eq(nil, cons(x0, x1))
eq(cons(x0, x1), cons(x2, x3))
eq(var(x0), var(x1))
eq(var(x0), apply(x1, x2))
eq(var(x0), lambda(x1, x2))
eq(apply(x0, x1), var(x2))
eq(apply(x0, x1), apply(x2, x3))
eq(apply(x0, x1), lambda(x2, x0))
eq(lambda(x0, x1), var(x2))
eq(lambda(x0, x1), apply(x1, x2))
eq(lambda(x0, x1), lambda(x2, x3))
if(true, var(x0), var(x1))
if(false, var(x0), var(x1))
ren(var(x0), var(x1), var(x2))
ren(x0, x1, apply(x2, x3))
ren(x0, x1, lambda(x2, x3))

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

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

Lexicographic Path Order [LPO].
Precedence:
var1 > EQ1

The following usable rules [FROCOS05] were oriented: none

(11) 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)))

The set Q consists of the following terms:

and(true, x0)
and(false, x0)
eq(nil, nil)
eq(cons(x0, x1), nil)
eq(nil, cons(x0, x1))
eq(cons(x0, x1), cons(x2, x3))
eq(var(x0), var(x1))
eq(var(x0), apply(x1, x2))
eq(var(x0), lambda(x1, x2))
eq(apply(x0, x1), var(x2))
eq(apply(x0, x1), apply(x2, x3))
eq(apply(x0, x1), lambda(x2, x0))
eq(lambda(x0, x1), var(x2))
eq(lambda(x0, x1), apply(x1, x2))
eq(lambda(x0, x1), lambda(x2, x3))
if(true, var(x0), var(x1))
if(false, var(x0), var(x1))
ren(var(x0), var(x1), var(x2))
ren(x0, x1, apply(x2, x3))
ren(x0, x1, lambda(x2, x3))

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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)))

The set Q consists of the following terms:

and(true, x0)
and(false, x0)
eq(nil, nil)
eq(cons(x0, x1), nil)
eq(nil, cons(x0, x1))
eq(cons(x0, x1), cons(x2, x3))
eq(var(x0), var(x1))
eq(var(x0), apply(x1, x2))
eq(var(x0), lambda(x1, x2))
eq(apply(x0, x1), var(x2))
eq(apply(x0, x1), apply(x2, x3))
eq(apply(x0, x1), lambda(x2, x0))
eq(lambda(x0, x1), var(x2))
eq(lambda(x0, x1), apply(x1, x2))
eq(lambda(x0, x1), lambda(x2, x3))
if(true, var(x0), var(x1))
if(false, var(x0), var(x1))
ren(var(x0), var(x1), var(x2))
ren(x0, x1, apply(x2, x3))
ren(x0, x1, lambda(x2, x3))

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

(15) 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: Combined order from the following AFS and order.
REN(x1, x2, x3)  =  REN(x3)
apply(x1, x2)  =  apply(x1, x2)
lambda(x1, x2)  =  lambda(x2)
ren(x1, x2, x3)  =  x3
var(x1)  =  var
cons(x1, x2)  =  cons(x2)
nil  =  nil
if(x1, x2, x3)  =  x3
eq(x1, x2)  =  x2
true  =  true
false  =  false
and(x1, x2)  =  and(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
REN1 > nil
apply2 > nil
lambda1 > nil
cons1 > nil
true > nil
false > var > nil
and2 > nil

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')

(16) 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)))

The set Q consists of the following terms:

and(true, x0)
and(false, x0)
eq(nil, nil)
eq(cons(x0, x1), nil)
eq(nil, cons(x0, x1))
eq(cons(x0, x1), cons(x2, x3))
eq(var(x0), var(x1))
eq(var(x0), apply(x1, x2))
eq(var(x0), lambda(x1, x2))
eq(apply(x0, x1), var(x2))
eq(apply(x0, x1), apply(x2, x3))
eq(apply(x0, x1), lambda(x2, x0))
eq(lambda(x0, x1), var(x2))
eq(lambda(x0, x1), apply(x1, x2))
eq(lambda(x0, x1), lambda(x2, x3))
if(true, var(x0), var(x1))
if(false, var(x0), var(x1))
ren(var(x0), var(x1), var(x2))
ren(x0, x1, apply(x2, x3))
ren(x0, x1, lambda(x2, x3))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE