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Proof step 14 from proof square-isSquare (previous step)

i % 10 = 0, ∀ i1, i0. i→n(i1) < i→n(i) → true → isSquare(mkDecimal(i1 * i1, 2 * abs(i0)).int) ⊦ i = 0, i0 = 0, isSquare(mkDecimal((i * i) / 10, 2 * abs(i0) - 1).int)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 16:
10 ≠ 0, ∃ j0. i = j0 * 10, ∀ i1, i0. i→n(i1) < i→n(i) → true → isSquare(mkDecimal(i1 * i1, 2 * abs(i0)).int) ⊦ i = 0, i0 = 0, isSquare(mkDecimal((i * i) / 10, 2 * abs(i0) - 1).int)

Rule argument

spec: int-div inst: name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [10, i] flags: rewrite

Additional information

Used terms: 10, i