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Proof step 73 from proof int2bytes-len-ls-01 (previous step)

¬ i < 2 ^ (n→i(n) * 8 - 1), # int2bits(i) / 8 = i→n(i0), # int2bits(i) ≤ i→n(8 * i0) + 1, 0 < i, i→n(8 * i0 - 1) + 1 < # int2bits(i), 0 < i0 ⊦ n→i(n) ≤ i0, i0 = 0, i = 0, # int2bits(i) % 8 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 74:
# int2bits(i) / 8 = i→n(i0), 2 ^ (n→i(n) * 8 - 1) ≤ i, i0 < n→i(n), 0 < i0, i→n(8 * i0 - 1) + 1 < # int2bits(i), 0 < i, # int2bits(i) ≤ i→n(8 * i0) + 1 ⊦ # int2bits(i) % 8 = 0, i = 0, i0 = 0, n = 0, n = 1

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ 0 < n ↔ n ≠ 0;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;