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Proof step 4 from proof int2bits-len-range (previous step)

0 < abs(j) → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ abs(j) ∧ abs(j) < 2 ^ (i + 1)), i→n(i + 1) < # int2bits(j), 0 ≤ 2 ^ i + j, j < 2 ^ i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i→n(i + 1) < # int2bits(j), 0 ≤ 2 ^ i + j, j < 2 ^ i, j ≠ 0 → 0 ≤ i0 ∧ 2 ^ i0 ≤ abs(j) ∧ abs(j) < 2 ^ (i0 + 1) ⊦

Rule argument

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m < 1 ↔ m = 0;% ⊦ # boolvar ' = 0 +1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ 0 < abs(i) ↔ i ≠ 0;