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Proof step 18 from proof div2-nls-02 (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = i0 * 2 + 1 - 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%mods-eq: ⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ 0 ≤ j1 ∧ j0 < j ∧ j1 < j → (i * j + j0 = i0 * j + j1 ↔ i = i0 ∧ j0 = j1); 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-04: ⊦ i + j < (* i1) * i + j0 ↔ j < (i1 - 1) * i + j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used formulas: i = i0 * 2 + 1 - j