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Proof step 5 from proof int2bits-len-neg-01 (previous step)

0 < 2 ^ i0, 2 ^ i0 + i < 0, 0 ≤ i0, 0 ≤ 2 ^ i0 * 2 + i, i < i0, 2 ^ i0 + i < i0, i + 1 < 2 ^ i0, i < 2 ^ i0, i < ~1, i < 0 ⊦ # int2bits(i) = # int2bits(~(i + 1)), i = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # int2bits(j) = i→n(i + 2)
Proof step 7:
0 ≤ i0 ∧ 2 ^ i0 ≤ ~(i + 1) ∧ ~(i + 1) < 2 ^ (i0 + 1), 0 < 2 ^ i0, 2 ^ i0 + i < 0, 0 ≤ i0, 0 ≤ 2 ^ i0 * 2 + i, i < i0, 2 ^ i0 + i < i0, i + 1 < 2 ^ i0, i < 2 ^ i0, i < ~1, i < 0 ⊦ # int2bits(i) = i→n(i0 + 2), i = ~1

Rule argument

name: int2bits-len-pos rotate: F subst: [i, j] → [i0, ~(i + 1)]

Additional information

Used simplifier rules: ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ i * j = j * i; Used terms: i0, ~(i + 1)