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Proof step 69 from proof equality-any (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
mkDecimal(i, j) = mkDecimal(0, 0), ∀ i1, i2, j, j0. i→n(i1) + i→n(i2) < i→n(i) + i→n(0) → true → (mkDecimal(i1, j) = mkDecimal(i2, j0) ↔ i1 * 10 ^ abs(j0) = i2 * 10 ^ abs(j)) ⊦ i % 10 = 0, j = 0, j0 = 0, i = 0

Rule argument

Additional information

Used simplifier rules: absls: ⊦ ¬ abs(i) < 0; from specification int-abs%pot-zero-01: ⊦ 0 < i → (i ^ j = 0 ↔ j < 0 ∧ 1 < i); from specification int-pot