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Proof step 97 from proof uint2bits-add-high (previous step)

abs(j - 1) / 2 = 2 ^ i0, 2 ^ (i - 1) = 2 ^ i0, j % 2 = 0, 2 ^ i0 + (j - 1) / 2 = 0, 2 ^ i0 + j < i0, 2 ^ i0 + j < i, j + 1 < i0, 0 < i0, j < i0, 0 < 2 ^ i, 1 < 2 ^ i, 2 ^ i0 + 1 < 2 ^ i, 0 < j + 2 ^ i, j + 1 < i, 0 < i, j < i, j + 1 < 2 ^ i, j < 2 ^ i, 0 ≤ 2 ^ i, 0 ≤ j + 2 ^ i, j ≤ 2 ^ i, j < 0, 2 ^ i0 + j < 2 ^ i, 2 ^ i0 + j < 0, j < (j - 1) / 2, (2 ^ i0 + j) + 1 < i0, (2 ^ i0 + j) + 1 < 2 ^ i, (2 ^ i0 + j) + 1 < i, ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ 2 ^ (i0 - 1) < abs(j) ∧ abs(j) ≤ 2 ^ i0 → uint2bits(j + 2 ^ i0) = sgn(int2bits(j).rest) ⊦ i = 1, j = 0, j = ~1, j + 2 ^ i = 1, i0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ 0 ≤ i ∧ 0 ≤ j → (2 ^ i = 2 ^ j ↔ i = j) subst: [j, i] → [i0, i - 1] rotate: F spec: int-pot inst: name: eq-two

Additional information

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