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Proof step 9 from proof prime-fac-divides (previous step)

0 < j, i < j, prime(i), ∀ j0, i. i→n(j0) < i→n(j) → prime(i) ∧ i < j0 ∧ 0 < j0 → i | fac(j0) ⊦ i | j, i | fac(j - 1), i + 1 < j ∧ 1 < j

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: prime-fac-divides-01: ⊦ prime(i) → i | fac(i);%fac-zero: ⊦ i ≤ 0 → fac(i) = 1;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%sub-eq-01: ⊦ i = j → i - j = 0; from specification int-abs%div-one-one: ⊦ i | 1 ↔ i = 1 ∨ i = ~1; from specification int-gcd%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%prime-not-neg-01: ⊦ j ≤ 0 → ¬ prime(j); from specification int-gcd%les: ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); 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%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2