Term Rewriting System R:
[x, y, z]
del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
=(nil, nil) -> true
=(.(x, y), nil) -> false
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)
DEL(.(x, .(y, z))) -> ='(x, y)
F(true, x, y, z) -> DEL(.(y, z))
F(false, x, y, z) -> DEL(.(y, z))
='(.(x, y), .(u, v)) -> ='(x, u)
='(.(x, y), .(u, v)) -> ='(y, v)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

F(false, x, y, z) -> DEL(.(y, z))
F(true, x, y, z) -> DEL(.(y, z))
DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)


Rules:


del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
=(nil, nil) -> true
=(.(x, y), nil) -> false
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)
four new Dependency Pairs are created:

DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
DEL(.(.(x'', y''), .(.(u, v), z))) -> F(and(=(x'', u), =(y'', v)), .(x'', y''), .(u, v), z)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Instantiation Transformation


Dependency Pairs:

DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
F(true, x, y, z) -> DEL(.(y, z))
DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
F(false, x, y, z) -> DEL(.(y, z))


Rules:


del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
=(nil, nil) -> true
=(.(x, y), nil) -> false
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(true, x, y, z) -> DEL(.(y, z))
one new Dependency Pair is created:

F(true, nil, nil, z'') -> DEL(.(nil, z''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 3
                 ↳Instantiation Transformation


Dependency Pairs:

DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
F(true, nil, nil, z'') -> DEL(.(nil, z''))
DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
F(false, x, y, z) -> DEL(.(y, z))
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)


Rules:


del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
=(nil, nil) -> true
=(.(x, y), nil) -> false
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(false, x, y, z) -> DEL(.(y, z))
two new Dependency Pairs are created:

F(false, .(x'''', y''''), nil, z'') -> DEL(.(nil, z''))
F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 4
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
F(true, nil, nil, z'') -> DEL(.(nil, z''))
DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
F(false, .(x'''', y''''), nil, z'') -> DEL(.(nil, z''))
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)


Rules:


del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
=(nil, nil) -> true
=(.(x, y), nil) -> false
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes