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Proof step 13 from proof int2bytes-rec-neg-05 (previous step)

# int2bytes(j) ≥ i→n(i) + 1, # int2bytes(j) = i→n(i) + 1, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ j = 0, bits2bytes(int2bits(j)) = i2b(~1) ' + lastn(i→n(i) + 1, int2bytes(j)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# int2bytes(j) = i→n(i) + 1, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ j = 0, bits2bytes(int2bits(j)) = i2b(~1) ' + int2bytes(j).rest

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < j → abs(j) = j;% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1);% ⊦ m + 1 ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ n ≥ m ↔ m ≤ n;% ⊦ n = # bytes → lastn(n, bytes) = bytes;% ⊦ m + n = n + m;