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Proof step 10 from proof int2bits-lelen-05 (previous step)

2 ≤ n→i(m) ∧ 0 ≤ 2, i < ~1, i < 0, 2 ≤ m ⊦ m ≤ # int2bits(i) ↔ ¬ # int2bits(i) ≤ (i→n(n→i(m)) - i→n(2)) + 1, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i < ~1, i < 0, 2 ≤ m ⊦ m = 0, m ≤ # int2bits(i) ↔ ¬ # int2bits(i) ≤ (m - 2) + 1

Rule argument

Additional information

Used simplifier rules: num-n2i-ls-06: 0 ≤ i ⊦ (* i) ≤ n→i(n) ↔ i→n(i) ≤ n; from specification intnat%n2n: ⊦ i→n(n→i(n)) = n; from specification intnat