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Proof step 20 from proof div2-nls (previous step)

i + j = i0 * 2, j < i0, i ≤ j, 0 ≤ i0, i < i0 ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = i0 * 2 - j

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%sub-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0; from specification int-abs%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%numle-10: ⊦ 0 < i → ((* i) * j ≤ (* i) * j0 ↔ j ≤ j0); from specification int-mult%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used formulas: i = i0 * 2 - j