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Proof step 41 from proof int2bytes-last-01 (previous step)

i % 256 ≤ 255, 0 < i % 256, 127 < i % 256, 0 ≤ i, 0 < i, 256 ≤ i ⊦ ~256 = i, i2b(i % 256) = i2b(i)

User applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i2b(i * 256 + j) = i2b(j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;%⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0); (elim) Used terms: i, 256, i0, j