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Proof step 5 from proof int2bits-lslen (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
m = m0 + 1 ∧ m ≠ 0, ¬ m ≤ # int2bits(i) ↔ i < 2 ^ (n→i(m0) - 1), 0 < i ⊦ # int2bits(i) < m ↔ ¬ 2 ^ (n→i(m) - 2) ≤ i

Rule argument

seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m0] rotate: T spec: nat inst: name: elim-pred-c

Additional information

Used simplifier rules: ⊦ 0 ≤ m;% ⊦ i < j → ¬ j < i;% ⊦ ¬ n < 0;% ⊦ ¬ i ≤ j ↔ j < i;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0