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Proof step 6 from proof product-prime-divides (previous step)

prime(i), prime(i), ∀ i. prime(i) → (i | Π ints0 ↔ (∃ j. j ∈ ints0 ∧ i | j)) ⊦ i | i0 ∨ i | Π ints0 ↔ (∃ j. (j = i0 ∨ j ∈ ints0) ∧ i | j)

User applied rule cut formula

And obtained the following new goals

Proof step 7:
i | i0, prime(i), prime(i), ∀ i. prime(i) → (i | Π ints0 ↔ (∃ j. j ∈ ints0 ∧ i | j)) ⊦ i | i0 ∨ i | Π ints0 ↔ (∃ j. (j = i0 ∨ j ∈ ints0) ∧ i | j)
Proof step 9:
prime(i), prime(i), ∀ i. prime(i) → (i | Π ints0 ↔ (∃ j. j ∈ ints0 ∧ i | j)) ⊦ i | i0, i | i0 ∨ i | Π ints0 ↔ (∃ j. (j = i0 ∨ j ∈ ints0) ∧ i | j)

Rule argument

fma: i | i0

Additional information

Used formulas: i | i0