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Proof step 70 from proof int2bytes-len-ls-01 (previous step)

i→n(i0) * 8 ≠ 0, # int2bits(i) / 8 = i→n(i0), # int2bits(i) = i→n(i0) * 8 + 1, # int2bits(i) ≤ 1 + (i→n(n→i(n) * 8) - 1), i→n(i0) * 8 - 1 < i→n(i0) * 8, i→n(i0) * 8 < # int2bits(i), 0 < i, 0 < i0, n ≤ i→n(i0), # int2bits(i) ≤ i→n(n→i(n) * 8), i→n(i0) * 8 < i→n(n→i(n) * 8), i→n(i0) * 8 < i→n(n→i(n) * 8) ⊦ # int2bits(i) % 8 = 0, i = 0, i0 = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;