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Proof step 9 from proof int2bits-lslen-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 < i, # int2bits(i) ≤ m, i→n(n→i(m) - 1) + 1 < # int2bits(i), i→n(n→i(m) - 1) + 1 < m ⊦

Rule argument

Additional information

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