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Proof step 26 from proof prime-product (previous step)

Π ints0 ≤ 4 ^ j, Π ints1 ≤ 4 ^ (j + 1), Π ints = Π ints1 * Π ints0, perm(ints, ints1 + ints0), primes(ints), 4 ^ (2 * j + 1) < Π ints, max(ints) ≤ 2 * j + 1, 0 ≤ j, 0 < j, primes(ints0), primes(ints1), ints1 = [] ∨ max(ints1) ≤ j + 1, ints0 = [] ∨ j + 1 < min(ints0), j + 2 ≤ min(ints0) ∧ max(ints0) ≤ j * 2 + 1 → Π ints0 ≤ 4 ^ j, ∀ i0, ints. i→n(i0) < i→n(j + j + 1) → primes(ints) ∧ max(ints) ≤ i0 ∧ ¬ dups(ints) → Π ints ≤ 4 ^ i0 ⊦ ints = [], dups(ints1), dups(ints0), dups(ints)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
Π ints = Π ints1 * Π ints0, Π ints0 ≤ 4 ^ j, Π ints1 ≤ 4 ^ (j + 1), perm(ints, ints1 + ints0), primes(ints), 4 ^ (2 * j + 1) < Π ints, max(ints) ≤ 2 * j + 1, 0 ≤ j, 0 < j, primes(ints0), primes(ints1), ints1 = [] ∨ max(ints1) ≤ j + 1, ints0 = [] ∨ j + 1 < min(ints0), ∀ i0, ints. i→n(i0) < i→n(j + j + 1) → primes(ints) ∧ max(ints) ≤ i0 ∧ ¬ dups(ints) → Π ints ≤ 4 ^ i0 ⊦ ints = [], dups(ints1), dups(ints0), dups(ints)

Proof step 26 from proof prime-product (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information