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 QDPOrderProof (⇔)
        ↳16 QDP
        ↳17 PisEmptyProof (⇔)
        ↳18 TRUE

(0) Obligation:

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

and(false, false) → false
and(true, false) → false
and(false, true) → false
and(true, true) → true
eq(nil, nil) → true
eq(cons(T, L), nil) → false
eq(nil, cons(T, L)) → false
eq(cons(T, L), cons(Tp, Lp)) → and(eq(T, Tp), eq(L, Lp))
eq(var(L), var(Lp)) → eq(L, Lp)
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(Tp, Sp)) → and(eq(T, Tp), eq(S, Sp))
eq(apply(T, S), lambda(X, Tp)) → false
eq(lambda(X, T), var(L)) → false
eq(lambda(X, T), apply(Tp, Sp)) → false
eq(lambda(X, T), lambda(Xp, Tp)) → and(eq(T, Tp), eq(X, Xp))
if(true, var(K), var(L)) → var(K)
if(false, var(K), var(L)) → var(L)
ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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(Tp, Lp)) → AND(eq(T, Tp), eq(L, Lp))
EQ(cons(T, L), cons(Tp, Lp)) → EQ(T, Tp)
EQ(cons(T, L), cons(Tp, Lp)) → EQ(L, Lp)
EQ(var(L), var(Lp)) → EQ(L, Lp)
EQ(apply(T, S), apply(Tp, Sp)) → AND(eq(T, Tp), eq(S, Sp))
EQ(apply(T, S), apply(Tp, Sp)) → EQ(T, Tp)
EQ(apply(T, S), apply(Tp, Sp)) → EQ(S, Sp)
EQ(lambda(X, T), lambda(Xp, Tp)) → AND(eq(T, Tp), eq(X, Xp))
EQ(lambda(X, T), lambda(Xp, Tp)) → EQ(T, Tp)
EQ(lambda(X, T), lambda(Xp, Tp)) → EQ(X, Xp)
REN(var(L), var(K), var(Lp)) → IF(eq(L, Lp), var(K), var(Lp))
REN(var(L), var(K), var(Lp)) → EQ(L, Lp)
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(false, false) → false
and(true, false) → false
and(false, true) → false
and(true, true) → true
eq(nil, nil) → true
eq(cons(T, L), nil) → false
eq(nil, cons(T, L)) → false
eq(cons(T, L), cons(Tp, Lp)) → and(eq(T, Tp), eq(L, Lp))
eq(var(L), var(Lp)) → eq(L, Lp)
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(Tp, Sp)) → and(eq(T, Tp), eq(S, Sp))
eq(apply(T, S), lambda(X, Tp)) → false
eq(lambda(X, T), var(L)) → false
eq(lambda(X, T), apply(Tp, Sp)) → false
eq(lambda(X, T), lambda(Xp, Tp)) → and(eq(T, Tp), eq(X, Xp))
if(true, var(K), var(L)) → var(K)
if(false, var(K), var(L)) → var(L)
ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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(Tp, Lp)) → EQ(L, Lp)
EQ(cons(T, L), cons(Tp, Lp)) → EQ(T, Tp)
EQ(var(L), var(Lp)) → EQ(L, Lp)
EQ(apply(T, S), apply(Tp, Sp)) → EQ(T, Tp)
EQ(apply(T, S), apply(Tp, Sp)) → EQ(S, Sp)
EQ(lambda(X, T), lambda(Xp, Tp)) → EQ(T, Tp)
EQ(lambda(X, T), lambda(Xp, Tp)) → EQ(X, Xp)

The TRS R consists of the following rules:

and(false, false) → false
and(true, false) → false
and(false, true) → false
and(true, true) → true
eq(nil, nil) → true
eq(cons(T, L), nil) → false
eq(nil, cons(T, L)) → false
eq(cons(T, L), cons(Tp, Lp)) → and(eq(T, Tp), eq(L, Lp))
eq(var(L), var(Lp)) → eq(L, Lp)
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(Tp, Sp)) → and(eq(T, Tp), eq(S, Sp))
eq(apply(T, S), lambda(X, Tp)) → false
eq(lambda(X, T), var(L)) → false
eq(lambda(X, T), apply(Tp, Sp)) → false
eq(lambda(X, T), lambda(Xp, Tp)) → and(eq(T, Tp), eq(X, Xp))
if(true, var(K), var(L)) → var(K)
if(false, var(K), var(L)) → var(L)
ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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(Tp, Lp)) → EQ(L, Lp)
EQ(cons(T, L), cons(Tp, Lp)) → EQ(T, Tp)
EQ(apply(T, S), apply(Tp, Sp)) → EQ(T, Tp)
EQ(apply(T, S), apply(Tp, Sp)) → EQ(S, Sp)
EQ(lambda(X, T), lambda(Xp, Tp)) → EQ(T, Tp)
EQ(lambda(X, T), lambda(Xp, Tp)) → EQ(X, Xp)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2)  =  EQ(x1, 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 with status [LPO].
Quasi-Precedence:
trivial

Status:
EQ2: [2,1]
cons2: [2,1]
apply2: [2,1]
lambda2: [2,1]


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(Lp)) → EQ(L, Lp)

The TRS R consists of the following rules:

and(false, false) → false
and(true, false) → false
and(false, true) → false
and(true, true) → true
eq(nil, nil) → true
eq(cons(T, L), nil) → false
eq(nil, cons(T, L)) → false
eq(cons(T, L), cons(Tp, Lp)) → and(eq(T, Tp), eq(L, Lp))
eq(var(L), var(Lp)) → eq(L, Lp)
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(Tp, Sp)) → and(eq(T, Tp), eq(S, Sp))
eq(apply(T, S), lambda(X, Tp)) → false
eq(lambda(X, T), var(L)) → false
eq(lambda(X, T), apply(Tp, Sp)) → false
eq(lambda(X, T), lambda(Xp, Tp)) → and(eq(T, Tp), eq(X, Xp))
if(true, var(K), var(L)) → var(K)
if(false, var(K), var(L)) → var(L)
ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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(Lp)) → EQ(L, Lp)
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 with status [LPO].
Quasi-Precedence:
trivial

Status:
EQ1: [1]
var1: [1]


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(false, false) → false
and(true, false) → false
and(false, true) → false
and(true, true) → true
eq(nil, nil) → true
eq(cons(T, L), nil) → false
eq(nil, cons(T, L)) → false
eq(cons(T, L), cons(Tp, Lp)) → and(eq(T, Tp), eq(L, Lp))
eq(var(L), var(Lp)) → eq(L, Lp)
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(Tp, Sp)) → and(eq(T, Tp), eq(S, Sp))
eq(apply(T, S), lambda(X, Tp)) → false
eq(lambda(X, T), var(L)) → false
eq(lambda(X, T), apply(Tp, Sp)) → false
eq(lambda(X, T), lambda(Xp, Tp)) → and(eq(T, Tp), eq(X, Xp))
if(true, var(K), var(L)) → var(K)
if(false, var(K), var(L)) → var(L)
ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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(false, false) → false
and(true, false) → false
and(false, true) → false
and(true, true) → true
eq(nil, nil) → true
eq(cons(T, L), nil) → false
eq(nil, cons(T, L)) → false
eq(cons(T, L), cons(Tp, Lp)) → and(eq(T, Tp), eq(L, Lp))
eq(var(L), var(Lp)) → eq(L, Lp)
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(Tp, Sp)) → and(eq(T, Tp), eq(S, Sp))
eq(apply(T, S), lambda(X, Tp)) → false
eq(lambda(X, T), var(L)) → false
eq(lambda(X, T), apply(Tp, Sp)) → false
eq(lambda(X, T), lambda(Xp, Tp)) → and(eq(T, Tp), eq(X, Xp))
if(true, var(K), var(L)) → var(K)
if(false, var(K), var(L)) → var(L)
ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REN(x1, x2, x3)  =  x3
apply(x1, x2)  =  apply(x1, x2)
lambda(x1, x2)  =  x2
ren(x1, x2, x3)  =  x3
var(x1)  =  var
cons(x1, x2)  =  x1
nil  =  nil
if(x1, x2, x3)  =  x2
eq(x1, x2)  =  eq(x1, x2)
true  =  true
false  =  false
and(x1, x2)  =  and(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
apply2 > eq2
nil > true
and2 > true
and2 > false > var > eq2

Status:
apply2: [2,1]
var: []
nil: []
eq2: [2,1]
true: []
false: []
and2: [2,1]


The following usable rules [FROCOS05] were oriented:

ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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:
The TRS P consists of the following rules:

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(false, false) → false
and(true, false) → false
and(false, true) → false
and(true, true) → true
eq(nil, nil) → true
eq(cons(T, L), nil) → false
eq(nil, cons(T, L)) → false
eq(cons(T, L), cons(Tp, Lp)) → and(eq(T, Tp), eq(L, Lp))
eq(var(L), var(Lp)) → eq(L, Lp)
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(Tp, Sp)) → and(eq(T, Tp), eq(S, Sp))
eq(apply(T, S), lambda(X, Tp)) → false
eq(lambda(X, T), var(L)) → false
eq(lambda(X, T), apply(Tp, Sp)) → false
eq(lambda(X, T), lambda(Xp, Tp)) → and(eq(T, Tp), eq(X, Xp))
if(true, var(K), var(L)) → var(K)
if(false, var(K), var(L)) → var(L)
ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
lambda2 > [var, nil, true]
cons > [var, nil, true]
eq > [var, nil, true]
apply > [var, nil, true]
[false, and2] > [var, nil, true]

Status:
lambda2: [2,1]
var: []
cons: []
nil: []
eq: []
apply: []
true: []
false: []
and2: [2,1]


The following usable rules [FROCOS05] were oriented:

ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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(false, false) → false
and(true, false) → false
and(false, true) → false
and(true, true) → true
eq(nil, nil) → true
eq(cons(T, L), nil) → false
eq(nil, cons(T, L)) → false
eq(cons(T, L), cons(Tp, Lp)) → and(eq(T, Tp), eq(L, Lp))
eq(var(L), var(Lp)) → eq(L, Lp)
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(Tp, Sp)) → and(eq(T, Tp), eq(S, Sp))
eq(apply(T, S), lambda(X, Tp)) → false
eq(lambda(X, T), var(L)) → false
eq(lambda(X, T), apply(Tp, Sp)) → false
eq(lambda(X, T), lambda(Xp, Tp)) → and(eq(T, Tp), eq(X, Xp))
if(true, var(K), var(L)) → var(K)
if(false, var(K), var(L)) → var(L)
ren(var(L), var(K), var(Lp)) → if(eq(L, Lp), var(K), var(Lp))
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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE