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Proof step 5 from proof put-firstn-restn (previous step)

n ≤ # y ⊦ (a0 ' + y)[n, a] = firstn(n, a0 ' + y) + a ' + restn(n, y)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n)
Proof step 7:
# y0 = n, y = y0 + z, n ≤ # y ⊦ (a0 ' + y)[n, a] = firstn(n, a0 ' + y) + a ' + z

Rule argument

spec: <toplevel> name: elim-firstn-restn seq: ⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n) subst: [x, n, y, z] → [y, n, y0, z] flags: <rotated> <all paths>

Additional information

Used terms: y, n, y0, z