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Proof step 7 from proof div2-nle-02 (previous step)

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

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-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%zero-ge-add-01: ⊦ 0 ≤ j ∧ 0 ≤ j0 ∧ 0 ≤ i0 → 0 ≤ i0 * j + j0; from specification int-mult%numls-03: ⊦ (* i2) * i + j < i + j0 ↔ (i2 - 1) * i + j < j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2 Used formulas: i = i0 * 2 + 1 - j