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Proof step 3 from proof square-eq (previous step)

abs(i) < abs(j), i * i = j * j ⊦ abs(i) = abs(j)

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: pls-pos seq: 0 < i, 0 < j, i ≤ i0, j < j0 ⊦ i * j < i0 * j0 subst: [i, j, i0, j0] → [abs(i), abs(i), abs(j), abs(j)] flags: current, []

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%square-zero: ⊦ i * i = 0 ↔ i = 0;%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs Used terms: abs(i), abs(i), abs(j), abs(j)