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Proof step 5 from proof prime-three (previous step)

3 % i = 0, i * i ≤ 3, 0 < i, 1 < i ⊦ 2 * 2 ≤ i * i

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: square-le seq: ⊦ i * i ≤ j * j ↔ abs(i) ≤ abs(j) subst: [j, i] → [i, 2] flags: rewrite

Additional information

Used simplifier rules: abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%two-ls: ⊦ 0 < j → (j < 2 ↔ j = 1); from specification int-basic1%one-01: ⊦ i = 1 → j % i = 0; from specification int-div%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used terms: i, 2