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Proof step 14 from proof div2-nls-02 (previous step)

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

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-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%numls-08: ⊦ i + j < (* i1) * i ↔ j < (i1 - 1) * i; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used formulas: i = i0 * 2 - j