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Proof step 11 from proof int2bytes-first-zerobyte-01 (previous step)

0 ≤ i0 * 8 + j0, # int2bits(j) = i→n(i0 * 8 + j0), j0 < 8, 0 ≤ j, 0 ≤ j0 ⊦ 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 12:
# int2bits(j) = i→n(i0 * 8 + j0), j0 < 8, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦ 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 = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);