Termination w.r.t. Q proof of /tmp/tmpOoRZ6p/#4.35.xml

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 PisEmptyProof (⇔)

        ↳9 TRUE

    ↳10 QDP

        ↳11 QDPOrderProof (⇔)

        ↳12 QDP

        ↳13 PisEmptyProof (⇔)

        ↳14 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(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 remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(EQ(x₁, x₂)) = x₂
POL(apply(x₁, x₂)) = 1 + x₁ + x₂
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(lambda(x₁, x₂)) = 1 + x₁ + x₂
POL(var(x₁)) = 1 + x₁

The following usable rules [FROCOS05] were oriented: none

(7) 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.

(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:

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.

(11) 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: Polynomial interpretation [POLO]:

POL(REN(x₁, x₂, x₃)) = x₃
POL(and(x₁, x₂)) = x₂
POL(apply(x₁, x₂)) = 1 + x₁ + x₂
POL(cons(x₁, x₂)) = 0
POL(eq(x₁, x₂)) = 0
POL(false) = 0
POL(if(x₁, x₂, x₃)) = 0
POL(lambda(x₁, x₂)) = 1 + x₂
POL(nil) = 0
POL(ren(x₁, x₂, x₃)) = x₃
POL(true) = 0
POL(var(x₁)) = 0

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')

(12) 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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) PisEmptyProof (EQUIVALENT transformation)

(9) TRUE

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) PisEmptyProof (EQUIVALENT transformation)

(14) TRUE