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Proof step 3 from proof square-eq-02 (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

Proof step 4:
⊦ 0 ≤ i ∧ 0 ≤ j → (i * (i + 1) = j * (j + 1) ↔ i = j)

Rule argument

spec: int-mult inst: name: square-eq-01 seq: ⊦ 0 ≤ i ∧ 0 ≤ j → (i * (i + 1) = j * (j + 1) ↔ i = j) subst: [j, i] → [j + 1, i] flags: rewrite, current

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%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 Used terms: j + 1, i