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Proof step 38 from proof equality-any (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
∀ i1, i2, j, j1. i→n(i1) + i→n(i2) < i→n(i) + i→n(i * 10 ^ abs(j0)) → true → (mkDecimal(i1, j) = mkDecimal(i2, j1) ↔ i1 * 10 ^ abs(j1) = i2 * 10 ^ abs(j)) ⊦ j0 = 0, (i * 10 ^ abs(j0)) % 10 = 0, i = 0, 10 ^ abs(j0) % 10 = 0

Proof step 38 from proof equality-any (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information