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Proof step 29 from proof int2bytes-first-neg (previous step)

int2bytes(i1 / 256) = by ' + bytes0, i1 < 0, i1 < ~383, 0 ≤ b2i(by), i1 / 256 ≤ 255 + i1, ∀ bytes, by, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

User applied rule insert elim lemma

And obtained the following new goals

Proof step 30:
i1 = i * 256 + j ∧ j ≤ 0 ∧ abs(j) < 256 ∧ i ≤ 0 ∧ 0 < 256, int2bytes(i) = by ' + bytes0, i1 < 0, i1 < ~383, 0 ≤ b2i(by), i ≤ 255 + i1, ∀ bytes, by, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

Rule argument

seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [i1, 256, i, j] rotate: T spec: int-div inst: name: div-elim-negpos

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);%⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0); (elim) Used terms: i1, 256, i, j