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Proof step 5 from proof all_prime_divisors-in-01 (previous step)

¬(2 ≤ i ∧ ¬ prime(i)), ∀ i0, j. i→n(i0) < i→n(i) → true → (j ∈ all_prime_divisors(i0) ↔ prime(j) ∧ j | i0 ∧ 2 ≤ i0) ⊦ j ∈ all_prime_divisors(i) ↔ prime(j) ∧ j | i ∧ 2 ≤ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
prime(i), 2 ≤ i, prime(j), j | i, 0 < i, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → true → (j ∈ all_prime_divisors(i0) ↔ prime(j) ∧ j | i0 ∧ 2 ≤ i0) ⊦ j = i

Rule argument

Additional information

Used simplifier rules: all_prime_divisors-base: ⊦ i < 2 → all_prime_divisors(i) = [];%all_prime_divisors-prime: ⊦ prime(i) → all_prime_divisors(i) = i ';%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%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%div-zero: ⊦ i | i; from specification int-gcd%in-17: ⊦ i ∈ i0 ' ↔ i = i0; from specification list[intlista]%in-empty: ⊦ ¬ i ∈ []; from specification list[intlista]%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs