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Proof step 9 from proof int2bits-lslen (previous step)

n→i(m0) = i0 + 1, ¬ m0 < # int2bits(i) ↔ i < 2 ^ i0, 0 < i ⊦ # int2bits(i) ≤ m0 ↔ ¬ 2 ^ i0 ≤ i

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

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