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Proof step 25 from proof int2bytes-first-neg-01 (previous step)

int2bytes(i1 / 256) = i2b(0) ' + bytes0, i1 < 0, i1 < ~383, i2b(0) ∈ int2bytes(i1 / 256), ∀ bytes, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦

User applied rule apply induction

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: 0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: bytes0, i1 / 256