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Proof step 5 from proof int2bytes-first-zerobyte-01 (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n→i(# int2bits(j)) = i0 * 8 + j0, j0 < 8, 0 ≤ j, 0 ≤ j0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 0 < n ↔ n ≠ 0;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;