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Proof step 22 from proof prime-product (previous step)

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 ⊦ i→n(j + 1) < i→n(j + j + 1), ints = [], dups(ints1), dups(ints0), dups(ints)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
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), i→n(2 * j + 1) ≤ i→n(j) + 1, 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)

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%abs-succ-03: ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1; from specification intnat