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Proof step 10 from proof prime-pot-eq (previous step)

i ^ (i0 - 1) * i * j = i * i ^ (i0 - 1) * j, 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 - 1) * i * j = i ^ (i1 - 1) * i * j0 ↔ i ^ (i0 - 1) * i = i ^ (i1 - 1) * i ∧ j = j0, i | j0, i | j

User applied rule insert equation

And obtained the following new goals

Rule argument

pos: left 1 flags: discard

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