Term Rewriting System R:
[T, L, Tp, Lp, S, X, Sp, Xp, K, Y, Z]
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)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pairs:

EQ(lambda(X, T), lambda(Xp, Tp)) -> EQ(X, Xp)
EQ(lambda(X, T), lambda(Xp, Tp)) -> EQ(T, Tp)
EQ(apply(T, S), apply(Tp, Sp)) -> EQ(S, Sp)
EQ(apply(T, S), apply(Tp, Sp)) -> EQ(T, Tp)
EQ(var(L), var(Lp)) -> EQ(L, Lp)
EQ(cons(T, L), cons(Tp, Lp)) -> EQ(L, Lp)
EQ(cons(T, L), cons(Tp, Lp)) -> EQ(T, Tp)


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)))





The following dependency pair can be strictly oriented:

EQ(var(L), var(Lp)) -> EQ(L, Lp)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(apply(x1, x2))=  x1 + x2  
  POL(EQ(x1, x2))=  x1  
  POL(var(x1))=  1 + x1  
  POL(cons(x1, x2))=  x1 + x2  
  POL(lambda(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pairs:

EQ(lambda(X, T), lambda(Xp, Tp)) -> EQ(X, Xp)
EQ(lambda(X, T), lambda(Xp, Tp)) -> EQ(T, Tp)
EQ(apply(T, S), apply(Tp, Sp)) -> EQ(S, Sp)
EQ(apply(T, S), apply(Tp, Sp)) -> EQ(T, Tp)
EQ(cons(T, L), cons(Tp, Lp)) -> EQ(L, Lp)
EQ(cons(T, L), cons(Tp, Lp)) -> EQ(T, Tp)


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)))





The following dependency pairs can be strictly oriented:

EQ(cons(T, L), cons(Tp, Lp)) -> EQ(L, Lp)
EQ(cons(T, L), cons(Tp, Lp)) -> EQ(T, Tp)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(apply(x1, x2))=  x1 + x2  
  POL(EQ(x1, x2))=  x1  
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(lambda(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Polo
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pairs:

EQ(lambda(X, T), lambda(Xp, Tp)) -> EQ(X, Xp)
EQ(lambda(X, T), lambda(Xp, Tp)) -> EQ(T, Tp)
EQ(apply(T, S), apply(Tp, Sp)) -> EQ(S, Sp)
EQ(apply(T, S), apply(Tp, Sp)) -> EQ(T, Tp)


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)))





The following dependency pairs can be strictly oriented:

EQ(lambda(X, T), lambda(Xp, Tp)) -> EQ(X, Xp)
EQ(lambda(X, T), lambda(Xp, Tp)) -> EQ(T, Tp)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(apply(x1, x2))=  x1 + x2  
  POL(EQ(x1, x2))=  x1  
  POL(lambda(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Polo
             ...
               →DP Problem 5
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pairs:

EQ(apply(T, S), apply(Tp, Sp)) -> EQ(S, Sp)
EQ(apply(T, S), apply(Tp, Sp)) -> EQ(T, Tp)


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)))





The following dependency pairs can be strictly oriented:

EQ(apply(T, S), apply(Tp, Sp)) -> EQ(S, Sp)
EQ(apply(T, S), apply(Tp, Sp)) -> EQ(T, Tp)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(apply(x1, x2))=  1 + x1 + x2  
  POL(EQ(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Polo
             ...
               →DP Problem 6
                 ↳Dependency Graph
       →DP Problem 2
         ↳Polo


Dependency Pair:


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)))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering


Dependency Pairs:

REN(X, Y, lambda(Z, T)) -> REN(Z, var(cons(X, cons(Y, cons(lambda(Z, T), nil)))), 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, apply(T, S)) -> REN(X, Y, S)
REN(X, Y, apply(T, S)) -> REN(X, Y, T)


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)))





The following dependency pairs can be strictly oriented:

REN(X, Y, apply(T, S)) -> REN(X, Y, S)
REN(X, Y, apply(T, S)) -> REN(X, Y, T)


Additionally, the following usable rules w.r.t. to the implicit AFS can be 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)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(REN(x1, x2, x3))=  x3  
  POL(and(x1, x2))=  0  
  POL(if(x1, x2, x3))=  0  
  POL(apply(x1, x2))=  1 + x1 + x2  
  POL(var(x1))=  0  
  POL(eq(x1, x2))=  0  
  POL(cons(x1, x2))=  0  
  POL(false)=  0  
  POL(lambda(x1, x2))=  x2  
  POL(nil)=  0  
  POL(true)=  0  
  POL(ren(x1, x2, x3))=  x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 7
             ↳Polynomial Ordering


Dependency Pairs:

REN(X, Y, lambda(Z, T)) -> REN(Z, var(cons(X, cons(Y, cons(lambda(Z, T), nil)))), T)
REN(X, Y, lambda(Z, T)) -> REN(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, T), nil)))), T))


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)))





The following dependency pairs can be strictly oriented:

REN(X, Y, lambda(Z, T)) -> REN(Z, var(cons(X, cons(Y, cons(lambda(Z, T), nil)))), T)
REN(X, Y, lambda(Z, T)) -> REN(X, Y, ren(Z, var(cons(X, cons(Y, cons(lambda(Z, T), nil)))), T))


Additionally, the following usable rules w.r.t. to the implicit AFS can be 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)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(REN(x1, x2, x3))=  x3  
  POL(and(x1, x2))=  0  
  POL(if(x1, x2, x3))=  0  
  POL(apply(x1, x2))=  0  
  POL(var(x1))=  0  
  POL(eq(x1, x2))=  0  
  POL(cons(x1, x2))=  0  
  POL(false)=  0  
  POL(lambda(x1, x2))=  1 + x2  
  POL(nil)=  0  
  POL(true)=  0  
  POL(ren(x1, x2, x3))=  x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 7
             ↳Polo
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:


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)))





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes