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Proof step 4 from proof square-le-05 (previous step)

0 ≤ i, 0 ≤ j ⊦ i * (i + 1) ≤ (j + 1) * ((j + 1) + 1) ↔ i ≤ j + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: square-le-04 seq: ⊦ 0 ≤ i ∧ 0 ≤ j → (i * (i + 1) ≤ j * (j + 1) ↔ i ≤ j) subst: [j, i] → [j + 1, i] flags: <all paths>

Additional information

Used simplifier rules: dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%numsucc: ⊦ (* i0) * i + i = (i0 + 1) * i;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; (s) Used terms: j + 1, i