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Proof step 18 from proof square-interval (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: zero-eq seq: ⊦ 0 ≤ i ∧ 0 ≤ j → (i + j = 0 ↔ i = 0 ∧ j = 0) subst: [j, i] → [j0, (j * (j + 1)) / 2] flags: <all paths>

Additional information

Used simplifier rules: div-two-zero-01: ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;%div-two-zero-01: ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ 0 + i = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; (s) Used terms: j0, (j * (j + 1)) / 2