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Proof step 12 from proof int2bits-lslen-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
0 < i, # int2bits(i) ≤ m, 1 + (m - 1) < # int2bits(i), 1 + (m - 1) < m ⊦ m = 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ n -1 = n - 1;% ⊦ i→n(n→i(n)) = n;% ⊦ m < n → n ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;