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Proof step 7 from proof int2bits-lslen (previous step)

¬ m0 < # int2bits(i) ↔ i < 2 ^ (n→i(m0) - 1), 0 < i ⊦ # int2bits(i) ≤ m0 ↔ ¬ 2 ^ (n→i(m0) - 1) ≤ i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
n→i(m0) = i0 + 1, ¬ m0 < # int2bits(i) ↔ i < 2 ^ i0, 0 < i ⊦ # int2bits(i) ≤ m0 ↔ ¬ 2 ^ i0 ≤ i

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [n→i(m0), 1, i0] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: n→i(m0), 1, i0