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Proof step 18 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 insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [i, j, i0, j0] → [i, 2, i1, j0] rotate: T spec: int-div inst: name: div-elim-pos

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i % 2 = 1 ∧ 0 < i → i / 2 * 2 + 1 = i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;%⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j); (elim) Used terms: i, 2, i1, j0