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Proof step 6 from proof int2bits-lslen-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 < i, m < # int2bits(i), # int2bits(i) ≤ i→n(n→i(m) - 1) + 1, m ≤ i→n(n→i(m) - 1) ⊦ m = 0

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)