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Proof step 2 from proof exodd (previous step)

i % 2 = 1, 0 ≤ i ⊦ ∃ j. 0 ≤ j ∧ i = 2 * j + 1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j)

Rule argument

spec: <toplevel> name: div-elim-pos 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, i0, j0] flags: <rotated> <all paths>

Additional information

Used simplifier rules: nummult-eq-01: j ≠ 0, j0 % j = 0 ⊦ i * (* j) = i0 * (* j0) ↔ i = i0 * j0 / j;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ 1 * i = i; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) Used terms: i, 2, i0, j0