Term Rewriting System R:
[y, t, l, t', l', s, x, s', x', k, z]
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)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

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


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 following dependency pair can be strictly oriented:

EQ(var(l), var(l')) -> EQ(l, l')


There are no usable rules using the Ce-refinement 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(x', t')) -> EQ(t, t')
EQ(lambda(x, t), lambda(x', t')) -> EQ(x, x')
EQ(apply(t, s), apply(t', s')) -> EQ(s, s')
EQ(apply(t, s), apply(t', s')) -> EQ(t, t')
EQ(cons(t, l), cons(t', l')) -> EQ(l, l')
EQ(cons(t, l), cons(t', l')) -> EQ(t, t')


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 following dependency pairs can be strictly oriented:

EQ(cons(t, l), cons(t', l')) -> EQ(l, l')
EQ(cons(t, l), cons(t', l')) -> EQ(t, t')


There are no usable rules using the Ce-refinement 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(x', t')) -> EQ(t, t')
EQ(lambda(x, t), lambda(x', t')) -> EQ(x, x')
EQ(apply(t, s), apply(t', s')) -> EQ(s, s')
EQ(apply(t, s), apply(t', s')) -> EQ(t, t')


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 following dependency pairs can be strictly oriented:

EQ(lambda(x, t), lambda(x', t')) -> EQ(t, t')
EQ(lambda(x, t), lambda(x', t')) -> EQ(x, x')


There are no usable rules using the Ce-refinement 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(t', s')) -> EQ(s, s')
EQ(apply(t, s), apply(t', s')) -> EQ(t, t')


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 following dependency pairs can be strictly oriented:

EQ(apply(t, s), apply(t', s')) -> EQ(s, s')
EQ(apply(t, s), apply(t', s')) -> EQ(t, t')


There are no usable rules using the Ce-refinement 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(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)))





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(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 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 using the Ce-refinement can be 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')
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'))
and(true, y) -> y
and(false, y) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(REN(x1, x2, x3))=  x3  
  POL(and(x1, x2))=  x2  
  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(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 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 using the Ce-refinement can be 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')
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'))
and(true, y) -> y
and(false, y) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(REN(x1, x2, x3))=  x3  
  POL(and(x1, x2))=  x2  
  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(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)))





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes