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Proof step 63 from proof byte-mult-256 (previous step)

⊦ j % 256 < 0, j < 0, b2i(i2b(i * 256 + j)) = b2i(i2b(j)), i = 0, ¬ i < 0 ∧ ¬ j < 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 64:
j = i0 * 256 + j0 ∧ 0 ≤ j0 ∧ j0 < 256 ∧ 0 ≤ i0 ∧ 0 < 256 ∧ 0 ≤ j ⊦ j0 < 0, j < 0, b2i(i2b(i * 256 + j)) = b2i(i2b(j)), i = 0, ¬ i < 0 ∧ ¬ j < 0

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] → [j, 256, i0, j0] rotate: T spec: int-div inst: name: div-elim-pospos

Additional information

Used simplifier rules: byte-equal: ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);%⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0); (elim) Used terms: j, 256, i0, j0