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Proof step 2 from proof one-02 (previous step)

true, ∀ j0. i→n(j0) < i→n(j) → true → 0 < j0 → 1 ^ j0 = 1 ⊦ 0 < j → 1 ^ j = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < j, ∀ j0. i→n(j0) < i→n(j) → true → 0 < j0 → 1 ^ j0 = 1 ⊦ 1 ^ j -1 = 1

Rule argument

Additional information

Used simplifier rules: Pot-pos: ⊦ 0 < i → j ^ i = j ^ i -1 * j;%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult