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Proof step 2 from proof prime-pot-eq (previous step)

true, ∀ i2, i, i1, j, j0. i→n(i2) < i→n(i0) → true → prime(i) ∧ 0 < i2 ∧ 0 < i1 ∧ ¬ i | j ∧ ¬ i | j0 → (i ^ i2 * j = i ^ i1 * j0 ↔ i ^ i2 = i ^ i1 ∧ j = j0) ⊦ prime(i) ∧ 0 < i0 ∧ 0 < i1 ∧ ¬ i | j ∧ ¬ i | j0 → (i ^ i0 * j = i ^ i1 * j0 ↔ i ^ i0 = i ^ i1 ∧ j = j0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
prime(i), 0 < i0, 0 < i1, ∀ i2, i, i1, j, j0. i→n(i2) < i→n(i0) → true → prime(i) ∧ 0 < i2 ∧ 0 < i1 ∧ ¬ i | j ∧ ¬ i | j0 → (i ^ i2 * j = i ^ i1 * j0 ↔ i ^ i2 = i ^ i1 ∧ j = j0) ⊦ i ^ i0 * j = i ^ i1 * j0 ↔ i ^ i0 = i ^ i1 ∧ j = j0, i | j0, i | j

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs