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Proof step 10 from proof int2bits-len-pos-pred (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: int2bits-len-pos rotate: F subst: [i, j] → [i, j - 1]

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + 1 ≤ j ↔ i < j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < j → ¬ j < i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ i * j = j * i; Used terms: i, j - 1