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Proof step 4 from proof square-interval (previous step)

j * (j + 1) + 2 * j0 < (j + 1) * (j + 2), j * (j + 1) + 2 * j0 < (i + 1) * (i + 2), i * (i + 1) < (j + 1) * ((j + 1) + 1), 0 ≤ j0, 0 ≤ i, j0 ≤ j, i * (i + 1) ≤ j * (j + 1) + 2 * j0 ⊦ i = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 ≤ i ∧ 0 ≤ j → (i * (i + 1) < j * (j + 1) ↔ i < j)
Proof step 6:
i < j + 1, j * (j + 1) + 2 * j0 < (i + 1) * (i + 2), j * (j + 1) + 2 * j0 < (j + 1) * (j + 2), i * (i + 1) ≤ j * (j + 1) + 2 * j0, j0 ≤ j, 0 ≤ j0, 0 ≤ i, 0 ≤ j + 1 ⊦ i = j

Rule argument

spec: <toplevel> name: square-ls-02 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;%numls-04: ⊦ i + j < (* i1) * i + j0 ↔ j < (i1 - 1) * i + j0;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≤ j → (j < i ↔ false); (as) Used terms: j + 1, i