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))

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
         ↳Argument Filtering and Ordering


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))





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{F, .}

resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2, x3, x4) -> F(x4)
DEL(x1) -> x1
.(x1, x2) -> .(x2)


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


Dependency Pairs:

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





Using the Dependency Graph resulted in no new DP problems.

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