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Proof step 7 from proof int2bytes-last-01 (previous step)

0 ≤ i, 0 < i, 128 ≤ i, i < 256 ⊦ i2b(i - 256) = i2b(i)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 256, i0] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i2b(i + 256) = i2b(i);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 256, i0