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Proof step 6 from proof zero (previous step)

j = i0 +1, 0 < j, i ^ i0 * i = 0, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → true → 0 < j0 → (i ^ j0 = 0 ↔ i = 0) ⊦ i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i ^ i0 = 0, 0 ≤ i0, ∀ j0, i. i→n(j0) < i→n(i0 +1) → true → 0 < j0 → (i ^ j0 = 0 ↔ i = 0) ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult