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Proof step 20 from proof square-nle (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: j = 0

Additional information

Used simplifier rules: lstwo-01: ⊦ i ≠ 0 ∧ 0 ≤ i → (i < 2 ↔ i = 1); from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: j = 0