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Proof step 7 from proof prime-product (previous step)

odd(i), 0 < i, 2 < i, 4 ^ i < Π ints, max(ints) ≤ i, primes(ints), ∀ i0, ints. i→n(i0) < i→n(i) → primes(ints) ∧ max(ints) ≤ i0 ∧ ¬ dups(ints) → Π ints ≤ 4 ^ i0 ⊦ ints = [], dups(ints)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ odd(i) ↔ (∃ j. i = j + j + 1)
Proof step 9:
∃ j. i = j + j + 1, 0 < i, 2 < i, 4 ^ i < Π ints, max(ints) ≤ i, primes(ints), ∀ i0, ints. i→n(i0) < i→n(i) → primes(ints) ∧ max(ints) ≤ i0 ∧ ¬ dups(ints) → Π ints ≤ 4 ^ i0 ⊦ ints = [], dups(ints)

Rule argument

spec: int-min inst: name: odd-def seq: ⊦ odd(i) ↔ (∃ j. i = j + j + 1) subst: [i] → [i] flags: rewrite

Additional information

Used terms: i