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Proof step 4 from proof elim-lastn-butlastn (previous step)

# z ≤ #(y + z) ⊦ y = firstn(#(y + z) - # z, y + z) ∧ z = lastn(# z, y + z)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ z = lastn(# z, y + z)

Rule argument

Additional information

Used simplifier rules: firstN-len: ⊦ firstn(# x, x + y) = x;%eq-01: ⊦ m + n - m = n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list