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Proof step 42 from proof equality-any (previous step)

∀ 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)) ⊦ i % 10 = 0, i0 % 10 = 0, i = i0 ∧ abs(j) = abs(j0) ↔ i * 10 ^ abs(j0) = i0 * 10 ^ abs(j), i = 0, i0 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 43:
i * 10 ^ abs(j0) = i0 * 10 ^ abs(j), ∀ 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)) ⊦ i = i0 ∧ abs(j) = abs(j0), i % 10 = 0, i0 % 10 = 0, i = 0, i0 = 0

Rule argument

pos: right 3

Additional information

Used simplifier rules: eq-nonzero: ⊦ i ≠ 0 → (i * j = i * j0 ↔ j = j0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult