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Proof step 13 from proof square-interval (previous step)

(j * (j + 1)) % 2 ≠ 1 ∧ 0 < j * (j + 1), ((i + 1) * (i + 2)) % 2 ≠ 1 ∧ 0 < (i + 1) * (i + 2), j * (j + 1) + 2 * j0 < (i + 1) * (i + 2), 0 ≤ j0, 0 ≤ j, 0 ≤ i, j0 ≤ j, (j * (j + 1)) / 2 + j0 < ((i + 1) * (i + 2)) / 2, (i * (i + 1)) / 2 ≤ (j * (j + 1)) / 2 + j0 ⊦ i * (i + 1) ≤ j * (j + 1) + 2 * j0, i = j

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 2 * (i * (i + 1)) / 2 ≤ 2 * ((j * (j + 1)) / 2 + j0)

Additional information

Used simplifier rules: square-even-02: ⊦ ((i + 1) * (i + 2)) % 2 = 0;%square-even: ⊦ 0 ≤ i → (i * (i + 1)) % 2 = 0;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%zero-gr-02: ⊦ 0 ≤ i → (0 < i * j ↔ ¬ ¬(0 < j ∧ i ≠ 0)); from specification int-mult%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%le-pos-01: ⊦ 0 < i → (j * i ≤ i0 * i ↔ j ≤ i0); from specification int-mult%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs Used formulas: 2 * (i * (i + 1)) / 2 ≤ 2 * ((j * (j + 1)) / 2 + j0)