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Proof step 3 from proof prime-ex-max (previous step)

0 < i, 2 ≤ i, 0 ≤ i ⊦ prime(i), ∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 1 < i → (prime(i) ↔ (∀ j. prime(j) ∧ j * j ≤ i → ¬ j | i))
Proof step 5:
0 < i, 1 < i, 0 ≤ i, 2 ≤ i ⊦ ∀ j. prime(j) ∧ j * j ≤ i → ¬ j | i, ∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i)

Rule argument

spec: <toplevel> name: prime-def-prime seq: ⊦ 1 < i → (prime(i) ↔ (∀ j. prime(j) ∧ j * j ≤ i → ¬ j | i)) subst: [i] → [i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s) Used terms: i