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Proof step 30 from proof equality-any (previous step)

¬(10 ^ abs(j) % 10 = 0 ∧ 10 ≠ 0), ∀ i1, i2, j1, j0. i→n(i1) + i→n(i2) < i→n(i0 * 10 ^ abs(j)) + i→n(i0) → true → (mkDecimal(i1, j1) = mkDecimal(i2, j0) ↔ i1 * 10 ^ abs(j0) = i2 * 10 ^ abs(j1)) ⊦ i0 % 10 = 0, (i0 * 10 ^ abs(j)) % 10 = 0, j = 0, i0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
∀ i1, i2, j1, j0. i→n(i1) + i→n(i2) < i→n(i0 * 10 ^ abs(j)) + i→n(i0) → true → (mkDecimal(i1, j1) = mkDecimal(i2, j0) ↔ i1 * 10 ^ abs(j0) = i2 * 10 ^ abs(j1)) ⊦ i0 % 10 = 0, (i0 * 10 ^ abs(j)) % 10 = 0, j = 0, i0 = 0, 10 ^ abs(j) % 10 = 0

Proof step 30 from proof equality-any (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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