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Proof step 4 from proof int2bits-lslen-06 (previous step)

0 < i ∧ 0 ≤ n→i(m) - 1, 0 < i ⊦ m < # int2bits(i) ↔ ¬ # int2bits(i) ≤ i→n(n→i(m) - 1) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i ⊦ m < # int2bits(i) ↔ ¬ # int2bits(i) ≤ i→n(n→i(m) - 1) + 1, m = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ 0 < n ↔ n ≠ 0;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;