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)))

Innermost 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
Argument Filtering and Ordering
       →DP Problem 2
AFS


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)))


Strategy:

innermost




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')
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')


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
EQ(x1, x2) -> EQ(x1, x2)
cons(x1, x2) -> cons(x1, x2)
var(x1) -> var(x1)
lambda(x1, x2) -> lambda(x1, x2)
apply(x1, x2) -> apply(x1, x2)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 3
Dependency Graph
       →DP Problem 2
AFS


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)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Argument Filtering and 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)))


Strategy:

innermost




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)


The following usable rules for innermost w.r.t. to the AFS 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')


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
REN(x1, x2, x3) -> x3
lambda(x1, x2) -> x2
ren(x1, x2, x3) -> x3
apply(x1, x2) -> apply(x1, x2)
var(x1) -> var
if(x1, x2, x3) -> x2


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 4
Argument Filtering and 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)))


Strategy:

innermost




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))


The following usable rules for innermost w.r.t. to the AFS 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')


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
REN(x1, x2, x3) -> x3
lambda(x1, x2) -> lambda(x2)
ren(x1, x2, x3) -> x3
var(x1) -> var
if(x1, x2, x3) -> x2
apply(x1, x2) -> apply(x1, x2)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 4
AFS
             ...
               →DP Problem 5
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)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:03 minutes