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Proof step 1 from proof square-inject (conclusion)

i * (i + 1) + 2 * i0 = j * (j + 1) + 2 * j0, i0 ≤ i, j0 ≤ j, 0 ≤ i0, 0 ≤ j, 0 ≤ j0 ⊦ i = j ∧ i0 = j0

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
i * (i + 1) ≤ j * (j + 1) + 2 * j0, j * (j + 1) + 2 * j0 < (i + 1) * (i + 2), j0 ≤ j, 0 ≤ i, 0 ≤ j0 ⊦ i = j
Proof step 3:
i * (i + 1) + 2 * i0 = j * (j + 1) + 2 * j0, i0 ≤ i, j0 ≤ j, 0 ≤ i0, 0 ≤ j, 0 ≤ j0 ⊦ i * (i + 1) ≤ j * (j + 1) + 2 * j0 ∧ j * (j + 1) + 2 * j0 < (i + 1) * (i + 2) ∧ 0 ≤ i, i = j ∧ i0 = j0

Rule argument

spec: int-mult inst: name: square-interval seq: i * (i + 1) ≤ j * (j + 1) + 2 * j0, j * (j + 1) + 2 * j0 < (i + 1) * (i + 2), j0 ≤ j, 0 ≤ i, 0 ≤ j0 ⊦ i = j subst: [i, j, j0] → [i, j, j0] flags: current, []

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;%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2 Used terms: i, j, j0