R
↳Dependency Pair Analysis
PURGE(.(x, y)) -> PURGE(remove(x, y))
PURGE(.(x, y)) -> REMOVE(x, y)
REMOVE(x, .(y, z)) -> REMOVE(x, z)
R
↳DPs
→DP Problem 1
↳Argument Filtering and Ordering
→DP Problem 2
↳Nar
REMOVE(x, .(y, z)) -> REMOVE(x, z)
purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))
innermost
REMOVE(x, .(y, z)) -> REMOVE(x, z)
POL(REMOVE(x1, x2)) = x1 + x2 POL(.(x1, x2)) = 1 + x1 + x2
REMOVE(x1, x2) -> REMOVE(x1, x2)
.(x1, x2) -> .(x1, x2)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 3
↳Dependency Graph
→DP Problem 2
↳Nar
purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))
innermost
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳Narrowing Transformation
PURGE(.(x, y)) -> PURGE(remove(x, y))
purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))
innermost
two new Dependency Pairs are created:
PURGE(.(x, y)) -> PURGE(remove(x, y))
PURGE(.(x'', nil)) -> PURGE(nil)
PURGE(.(x'', .(y'', z'))) -> PURGE(if(=(x'', y''), remove(x'', z'), .(y'', remove(x'', z'))))