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Proof step 37 from proof int2bytes-first-zerobyte (previous step)

2 ^ 8 ≤ i ∧ 0 < 8, 0 < i, 0 ≤ i, 255 < i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ int2bytes(i / 256).first = i2b(0) ↔ n→i(#(int2bits(i / 2 ^ 8) + fillsgn(i→n(8 + 1), int2bits(i % 2 ^ 8)).rest)) % 8 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;