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Proof step 4 from proof int2bits-lslen-12 (previous step)

# int2bits(i) < i→n(j) + 2, 0 ≤ j, 0 < i, 2 ^ j < i ⊦ i = 2 ^ j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i → (# int2bits(i) < m ↔ i < 2 ^ (n→i(m) - 2))

Rule argument

name: int2bits-lslen rotate: F subst: [m, i] → [i→n(j) + 2, i]

Additional information

Used simplifier rules: ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ 0 ≤ i → n→i(i→n(i)) = i;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + 0 = i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i < j → ¬ j < i; Used terms: i→n(j) + 2, i