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Proof step 2 from proof zero (previous step)

0 < j ⊦ i ^ j = 0 ↔ i = 0

User applied rule induction

And obtained the following new goals

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

Rule argument

indvar: i→n(j) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j)