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

0 ≤ i ∧ 0 ≤ 2, n→i(m) = i + 1, 0 ≤ i, j < 2 ^ (8 * (i + 1)), 2 ^ (8 * (i + 1) - 1) ≤ j, 0 ≤ j, j < 2 ^ ((i + 1) * 8), 2 ^ ((i + 1) * 8 - 1) ≤ j ⊦ m = 0, m + 1 = i→n(i) + i→n(2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n);