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Proof step 61 from proof equality-any (previous step)

i0 % 10 = 0 ∧ j0 ≠ 0, ∀ i1, i2, j, j0. i→n(i1) + i→n(i2) < i→n(i) + i→n(i0) → true → (mkDecimal(i1, j) = mkDecimal(i2, j0) ↔ i1 * 10 ^ abs(j0) = i2 * 10 ^ abs(j)) ⊦ mkDecimal(i, j) = mkDecimal(i0 / 10, abs(j0) - 1) ↔ i * 10 ^ abs(j0) = i0 * 10 ^ abs(j), i % 10 = 0 ∧ j ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
i0 % 10 = 0, ∀ i1, i2, j, j0. i→n(i1) + i→n(i2) < i→n(i) + i→n(i0) → true → (mkDecimal(i1, j) = mkDecimal(i2, j0) ↔ i1 * 10 ^ abs(j0) = i2 * 10 ^ abs(j)) ⊦ j0 = 0, mkDecimal(i, j) = mkDecimal(i0 / 10, abs(j0) - 1) ↔ i * 10 ^ abs(j0) = i0 * 10 ^ abs(j), i % 10 = 0 ∧ j ≠ 0

Proof step 61 from proof equality-any (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information