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Proof step 15 from proof int2bytes-first-neg (previous step)

int2bytes((i - 255) / 256) + int2bytes(i % 256).last ' = by ' + bytes, i < ~128, 0 ≤ b2i(by), i < 0, ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦ int2bytes(i % 256).last ≠ by ∧ ¬ int2bytes(i % 256).last ∈ bytes, ¬ by ∈ int2bytes((i - 255) / 256) ∧ by ≠ int2bytes(i % 256).last

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
i = i1 + 255, int2bytes(i1 / 256) + int2bytes(i % 256).last ' = by ' + bytes, i < ~128, 0 ≤ b2i(by), i < 0, ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦ int2bytes(i % 256).last ≠ by ∧ ¬ int2bytes(i % 256).last ∈ bytes, ¬ by ∈ int2bytes(i1 / 256) ∧ by ≠ int2bytes(i % 256).last

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 255, i1] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 255, i1