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Proof step 7 from proof isSquare-mult (previous step)

abs(j0) * abs(j0) = abs(j) * abs(i) * abs(i), j0 * j0 = j * i * i ⊦ i = 0, isSquare(j)

User applied rule insert lemma

And obtained the following new goals

Proof step 8:
j0 * j0 = j * i * i, 0 < j0, 0 < j, 0 < i ⊦ i = 0, isSquare(j)
Proof step 9:
abs(i) = 0 ∨ isSquare(abs(j)), abs(j0) * abs(j0) = abs(j) * abs(i) * abs(i), j0 * j0 = j * i * i ⊦ i = 0, isSquare(j)

Rule argument

spec: inst: name: isSquare-mult-aux seq: j0 * j0 = j * i * i, 0 < j0, 0 < j, 0 < i ⊦ i = 0, isSquare(j) subst: [j0, j, i] → [abs(j0), abs(j), abs(i)] flags: current, []

Additional information

Used simplifier rules: abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%Mult-zero: ⊦ i * 0 = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%square-zero: ⊦ i * i = 0 ↔ i = 0; from specification int-mult%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs%isSquare-zero: ⊦ isSquare(0); from specification int-square%zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult Used terms: abs(j0), abs(j), abs(i)