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Proof step 15 from proof prime-ex-max (previous step)

prime(j1), j1 | i, j1 < i, j < i, prime(j0), j0 < i, j < j0, j0 | i, 0 < i, 0 ≤ i, 2 ≤ i, prime(j), j | i, j + 1 < i, ∀ j0. j1 < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i, ∀ i0, j0. i→n(i0 - j0) < i→n(i - j) → j0 < i0 ∧ j0 | i0 ∧ prime(j0) ∧ 2 ≤ i0 ∧ 0 ≤ i0 ∧ 0 < i0 → (∃ j. prime(j) ∧ j | i0 ∧ j < i0 ∧ (∀ j0. j < j0 ∧ j0 < i0 ∧ prime(j0) → ¬ j0 | i0)) ⊦ ∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: j1 abort: F discard: T computed: T

Additional information

Used terms: j1