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Proof step 14 from proof prime-ex-max (previous step)

j < i, 0 < i, j < j0, j0 < i, prime(j), prime(j0), j | i, j0 | i, 0 ≤ i, 2 ≤ i, ∀ i0, j0. i→n(i0 - j0) < i→n(i - j) → j0 < i0 ∧ 0 < i0 ∧ prime(j0) ∧ j0 | i0 ∧ 2 ≤ 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 apply induction

And obtained the following new goals

Rule argument

subst: [i0, j0] → [i, j0]

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); (s)% ⊦ j < i → abs(i - j) = i - j; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i + (j - i0) = i + j - i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < j - i0 ↔ i + i0 < j; (s)% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: i, j0