R
↳Dependency Pair Analysis
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)
R
↳DPs
→DP Problem 1
↳Polynomial Ordering
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)
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
DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)
POL(and(x1, x2)) = 0 POL(DEL(x1)) = x1 POL(v) = 0 POL(false) = 0 POL(=(x1, x2)) = 0 POL(nil) = 0 POL(true) = 0 POL(.(x1, x2)) = 1 + x2 POL(u) = 0 POL(F(x1, x2, x3, x4)) = 1 + x4
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Dependency Graph
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)
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