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Proof step 2 from proof int2bits-lslen-06 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < i ∧ 0 ≤ j → (i < 2 ^ j ↔ # int2bits(i) ≤ i→n(j) + 1)
Proof step 4:
0 < i ∧ 0 ≤ n→i(m) - 1, 0 < i ⊦ m < # int2bits(i) ↔ ¬ # int2bits(i) ≤ i→n(n→i(m) - 1) + 1

Rule argument

name: int2bits-lslen-01 rotate: F subst: [j, i] → [n→i(m) - 1, i]

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 1 ≤ i ↔ 0 < i;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 0 < n ↔ n ≠ 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ i < j → ¬ j < i; Used terms: n→i(m) - 1, i