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Proof step 6 from proof prime-ex-max (previous step)

1 < i, 0 < i, prime(j), j | i, j * j ≤ i, 2 ≤ i, 0 ≤ i ⊦ ∃ j. prime(j) ∧ j | i ∧ j < i ∧ (∀ j0. j < j0 ∧ j0 < i ∧ prime(j0) → ¬ j0 | i)

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: j < i

Additional information

Used simplifier rules: ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs% ⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1); (s)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i = 1 → i * j = j; (s)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) Used formulas: j < i