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Proof step 15 from proof prime-pot-eq (previous step)

i ≠ 0, i ≠ 0, 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) * j = j0 * i ^ (i1 - 1) ↔ i ^ (i0 - 1) = i ^ (i1 - 1) ∧ j = j0, i | j0, i | j

User applied rule apply induction

And obtained the following new goals

Proof step 16:
prime(i) ∧ 0 < i0 - 1 ∧ 0 < i1 - 1 ∧ ¬ i | j ∧ ¬ i | j0 → (i ^ (i0 - 1) * j = i ^ (i1 - 1) * j0 ↔ i ^ (i0 - 1) = i ^ (i1 - 1) ∧ j = j0), i ≠ 0, i ≠ 0, 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) * j = j0 * i ^ (i1 - 1) ↔ i ^ (i0 - 1) = i ^ (i1 - 1) ∧ j = j0, i | j0, i | j

Rule argument

subst: [i2, i, i1, j, j0] → [i0 - 1, i, i1 - 1, j, j0]

Additional information

Used simplifier rules: abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: i0 - 1, i, i1 - 1, j, j0