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Proof step 2 from proof square-nle-01 (previous step)

((j + 1) * (j + 2)) / 2 ≤ j, 0 ≤ j ⊦

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: square-nle seq: ⊦ 0 ≤ j → ((j * (j + 1)) / 2 ≤ j ↔ j < 2) subst: [j] → [j + 1] flags: current, []

Additional information

Used simplifier rules: numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%le-add-03: ⊦ i ≤ i0 ∧ j0 ≥ 0 → i ≤ i0 + j0; from specification int-abs%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%bet-01: ⊦ 0 ≤ i → (i < 1 ↔ i = 0); from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j + 1