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Proof step 13 from proof square-isSquare (previous step)

(i * i) % 10 = 0 ∧ 2 * abs(i0) ≠ 0 ∧ i * i ≠ 0, (i * 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, abs(2 * abs(i0)) - 1).int)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: abs-01: ⊦ abs(abs(i)) = abs(i); from specification int-abs%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs%two-zero: ⊦ 2 * i = 0 ↔ i = 0; from specification int-mult%square-zero: ⊦ i * i = 0 ↔ i = 0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%mod-square-10: ⊦ (i * i) % 10 = 0 ↔ i % 10 = 0; from specification int-gcd