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Proof step 1 from proof int2bits-len-pos-01 (conclusion)

⊦ 0 ≤ j ∧ j < 2 ^ i → (i→n(i) < # int2bits(j) ↔ # int2bits(j) = i→n(i) + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward)% ⊦ m = n + 1 → ¬ m ≤ n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ m < n ↔ n ≤ m;