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Proof step 7 from proof int2bits-len-neg-01 (previous step)

0 ≤ i0 ∧ 2 ^ i0 ≤ ~(i + 1) ∧ ~(i + 1) < 2 ^ (i0 + 1), 0 < 2 ^ i0, 2 ^ i0 + i < 0, 0 ≤ i0, 0 ≤ 2 ^ i0 * 2 + i, i < i0, 2 ^ i0 + i < i0, i + 1 < 2 ^ i0, i < 2 ^ i0, i < ~1, i < 0 ⊦ # int2bits(i) = i→n(i0 + 2), i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;