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Proof step 17 from proof int2bytes-leading-zeros (previous step)

i2b(i - 256) = i2b(0), 0 < i, i ≤ 255, 0 ≤ i, 127 < i, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 256, 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, 256, i1