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Proof step 35 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)) ⊦ j ≠ 0, i0 % 10 = 0, i = i0 ∧ abs(j) = abs(j0) ↔ i * 10 ^ abs(j0) = i0 * 10 ^ abs(j), i = 0, i0 = 0, i % 10 = 0 ∧ j ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
∀ 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

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%zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult%eq-same-01: ⊦ i ≠ 0 → (i * j = i ↔ j = 1); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs