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Proof step 3 from proof zero (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i ^ j = 0, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → 0 < j0 → (i ^ j0 = 0 ↔ i = 0) ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: zero-01: ⊦ 0 < j → 0 ^ j = 0;% ⊦ 1 ≤ i ↔ 0 < i; (s)%lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs