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Proof step 2 from proof BertrandsPostulate (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 3:
2 < i, 0 < i ⊦ ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i
Proof step 31:
0 < i ⊦ 2 < i, ∃ j. prime(j) ∧ i < j ∧ j ≤ 2 * i

Rule argument

fma: 2 < i

Additional information

Used formulas: 2 < i