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Proof step 6 from proof int2bits-lslen-12 (previous step)

¬ # int2bits(i) < i→n(j) + 2 ∧ i ≠ 2 ^ j, 0 < i, 0 ≤ j ⊦ 2 ^ j < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i→n(j) + 2 ≤ # int2bits(i), i < 2 ^ j, 0 ≤ j, 0 < i, 0 < j ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ 1 < 2 ^ i ↔ 0 < i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ i < j → ¬ j < i;% ⊦ ¬ i < j → (i ≠ j ↔ j < i);