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Proof step 2 from proof int2bits-len-pos-01 (previous step)

i→n(i) + 1 < # int2bits(j), 0 < 2 ^ i, i→n(i) < # int2bits(j), j < 2 ^ i, 0 ≤ j ⊦ # int2bits(j) = i→n(i) + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
~(2 ^ i) ≤ j, j < 2 ^ i ⊦ # int2bits(j) ≤ i→n(i + 1)
Proof step 4:
# int2bits(j) ≤ i→n(i + 1), i→n(i) + 1 < # int2bits(j), 0 < 2 ^ i, i→n(i) < # int2bits(j), j < 2 ^ i, 0 ≤ j ⊦ # int2bits(j) = i→n(i) + 1

Rule argument

name: int2bits-len-range subst: [i, j] → [i, j]

Additional information

Used simplifier rules: ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ i + j < j ↔ i < 0;% ⊦ ¬ 2 ^ i < 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: i, j