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Proof step 17 from proof uint2bits-inj (previous step)

i % 2 = 1 ↔ j % 2 = 1, i / 2 = j / 2, 0 ≤ i, 0 ≤ j, 0 < i, 0 < j, ∀ i0, j. i→n(i0) < i→n(i) → uint2bits(i0) = uint2bits(j) ∧ 0 ≤ i0 ∧ 0 ≤ j → i0 = j ⊦ j = 1, i = 1, i = j

User applied rule case distinction

And obtained the following new goals

Proof step 18:
i % 2 = 1, j % 2 = 1, i / 2 = j / 2, 0 ≤ i, 0 ≤ j, 0 < i, 0 < j, ∀ i0, j. i→n(i0) < i→n(i) → uint2bits(i0) = uint2bits(j) ∧ 0 ≤ i0 ∧ 0 ≤ j → i0 = j ⊦ j = 1, i = 1, i = j

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ i % 2 ≠ 1 ∧ 0 < i → i % 2 = 0; (forward)% ⊦ i % 2 = 0 → (i / 2 = j ↔ i = 2 * j);% ⊦ i % 2 ≠ 1 ∧ 0 < i → i / 2 * 2 = i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;