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Proof step 3 from proof square-nls-02 (previous step)

(j * (j + 1)) / 2 < i + j, 0 ≤ j, i ≤ 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 ≤ j ∧ i < 0 → ¬ (j * (j + 1)) / 2 < i + j

Rule argument

spec: <toplevel> name: square-nls-01 seq: ⊦ 0 ≤ j ∧ i < 0 → ¬ (j * (j + 1)) / 2 < i + j subst: [i, j] → [i, j] flags: <all paths>

Additional information

Used simplifier rules: square-nls: ⊦ 0 ≤ j → ¬ (j * (j + 1)) / 2 < j;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: i, j