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Proof step 13 from proof prime-ex-max (previous step)

j < i, prime(j0), j0 < i, j < j0, j0 | i, 0 < i, 0 ≤ i, 2 ≤ i, prime(j), j | i, j + 1 < 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 apply induction

And obtained the following new goals

Rule argument

subst: [i0, j0] → [i, j0]

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abssub-02: ⊦ j < i → abs(i - j) = i - j; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%ls-add-both: ⊦ i < j → i1 + i < i1 + j; from specification int-basic2 Used terms: i, j0