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Proof step 11 from proof int2bits-lelen-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
i→n(n→i(m) - 2) + 1 < m, i→n(n→i(m) - 2) + 1 < # int2bits(i), # int2bits(i) < m, 2 ≤ m, 0 < i, 2 + i→n(n→i(m) - 2) < m ⊦

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%l: ⊦ # bits = 0 ↔ bits = []; from specification list-data[bitlista]%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat