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Proof step 3 from proof prime-odd (previous step)

prime(j) ⊦ j % 2 = 1, j = 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → (i % 2 = 1 ↔ i % 2 ≠ 0)
Proof step 5:
j % 2 = 0, 0 < j, prime(j) ⊦ j = 2

Rule argument

spec: int-div name: mod-two-06 seq: ⊦ 0 < i → (i % 2 = 1 ↔ i % 2 ≠ 0) subst: [i] → [j] flags: <all paths>

Additional information

Used simplifier rules: prime-not-neg-01: ⊦ j ≤ 0 → ¬ prime(j);% ⊦ i < j ↔ ¬ j ≤ i; (ss) Used terms: j