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Proof step 4 from proof int2bytes-first-zerobyte-01 (previous step)

0 ≤ j, 0 ≤ j ⊦ n→i(# int2bits(j)) % 8 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
n→i(# int2bits(j)) = i0 * 8 + j0 ∧ 0 ≤ j0 ∧ j0 < 8 ∧ 0 < 8 ∧ 0 < n→i(# int2bits(j)), 0 ≤ j, 0 ≤ j ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

Rule argument

seq: ⊦ 0 < j ∧ 0 < i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [i, j, i0, j0] → [n→i(# int2bits(j)), 8, i0, j0] rotate: T spec: int-div inst: name: div-elim

Additional information

Used simplifier rules: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 0 < n ↔ n ≠ 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ int2bits(j) ≠ [];%⊦ 0 < j ∧ 0 < i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j); (elim) Used terms: n→i(# int2bits(j)), 8, i0, j0