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Proof step 24 from proof int2bytes-first-neg (previous step)

int2bytes(i1 / 256) = by ' + bytes0, 0 ≤ b2i(by), i1 < ~383, i1 < 0, ∀ 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 apply induction

And obtained the following new goals

Rule argument

subst: [bytes, by, i0] → [bytes0, by, i1 / 256]

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);%0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ by ∈ by0 ' + bytes ↔ by ≠ by0 ∧ ¬ by ∈ bytes;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1; Used terms: bytes0, by, i1 / 256