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Proof step 30 from proof int2bytes-len-ls-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
# int2bits(i) = 1, 2 ^ (n→i(n) * 8 - 1) ≤ i, 0 < i ⊦ n = 0, i = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ m ≤ 0 ↔ m = 0;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ i→n(i) = 0 ↔ i = 0;%0 ≠ m ⊦ (* m) * n = 0 ↔ n = 0;% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;