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Proof step 9 from proof prime-product (previous step)

∃ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
primes(ints), 4 ^ (2 * j + 1) < Π ints, max(ints) ≤ 2 * j + 1, 0 ≤ j, 0 < j, ∀ i0, ints. i→n(i0) < i→n(j + j + 1) → primes(ints) ∧ max(ints) ≤ i0 ∧ ¬ dups(ints) → Π ints ≤ 4 ^ i0 ⊦ ints = [], dups(ints)

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%nummult-ls-02: ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0); from specification int-div%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult