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Proof step 3 from proof BertrandsPostulate (previous step)

2 < i, 0 < i ⊦ ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
1 < i ⊦ ∃ ints, ints0, ints1, ints2. all_prime_divisors(binom_coef(2 * i, i)) = ints + ints0 + ints1 + ints2 ∧ (∀ j. j ∈ ints → j ≤ sqrt(2 * i)) ∧ (∀ j. j ∈ ints0 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3) ∧ (∀ j. j ∈ ints1 → (2 * i) / 3 < j ∧ j ≤ i) ∧ (∀ j. j ∈ ints2 → i < j ∧ j ≤ 2 * i)
Proof step 5:
∃ ints, ints0, ints1, ints2. all_prime_divisors(binom_coef(2 * i, i)) = ints + ints0 + ints1 + ints2 ∧ (∀ j. j ∈ ints → j ≤ sqrt(2 * i)) ∧ (∀ j. j ∈ ints0 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3) ∧ (∀ j. j ∈ ints1 → (2 * i) / 3 < j ∧ j ≤ i) ∧ (∀ j. j ∈ ints2 → i < j ∧ j ≤ 2 * i), 2 < i, 0 < i ⊦ ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i

Rule argument

spec: inst: name: bertrand-prime-factors seq: 1 < i ⊦ ∃ ints, ints0, ints1, ints2. all_prime_divisors(binom_coef(2 * i, i)) = ints + ints0 + ints1 + ints2 ∧ (∀ j. j ∈ ints → j ≤ sqrt(2 * i)) ∧ (∀ j. j ∈ ints0 → sqrt(2 * i) < j ∧ j ≤ (2 * i) / 3) ∧ (∀ j. j ∈ ints1 → (2 * i) / 3 < j ∧ j ≤ i) ∧ (∀ j. j ∈ ints2 → i < j ∧ j ≤ 2 * i) subst: [i] → [i] flags: current, []

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%prime-le-2: ⊦ prime(i) → (i ≤ 2 ↔ i = 2); from specification int-gcd%prime-two-01: ⊦ prime(2); from specification int-gcd%two-ls: ⊦ 0 < j → (j < 2 ↔ j = 1); from specification int-basic1 Used terms: i