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Proof step 10 from proof div2-nls (previous step)

i0 * 2 = (i + j) + 1, i < 0, i < i0, j < 0, i0 ≤ 0, i ≤ j, j < i0 ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = i0 * 2 - j + 1

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + 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%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%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: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%numls-05: ⊦ (* i2) * i < i + j0 ↔ (i2 - 1) * i < j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used formulas: i = i0 * 2 - j + 1