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Proof step 21 from proof int2bytes-first-zerobyte-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 22:
j = 0, # int2bits(j) = i→n(j0) + 8 * i→n(i0), 0 ≤ i0, 0 ≤ j0, 0 ≤ j, j0 < 8 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j), (j0 ≠ 0 ∨ i0 ≠ 0) ∧ j ≠ 0

Rule argument

fma: j = 0

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ int2bits(j) ≠ []; Used formulas: j = 0