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Proof step 3 from proof minus-one-odd (previous step)

j % 2 = 1, 0 < j ⊦ ~1 ^ j = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i % 2 = 1 ↔ (∃ j. i = 2 * j + 1 ∧ 0 ≤ j)
Proof step 5:
0 < j, ∃ j0. j = 2 * j0 + 1 ∧ 0 ≤ j0 ⊦ ~1 ^ j = ~1

Rule argument

spec: int-div name: modex-odd seq: ⊦ i % 2 = 1 ↔ (∃ j. i = 2 * j + 1 ∧ 0 ≤ j) subst: [i] → [j] flags: <all paths>

Additional information

Used terms: j