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Proof step 8 from proof div2-nle-02 (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = i0 * 2 + 1 - j

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + (j - i) = j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ i + 0 = i; (s)% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i + i = 2 * i; (s)% ⊦ (* i2) * i + j < i + j0 ↔ (i2 - 1) * i + j < j0; (s)% ⊦ 1 * i = i; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i + j < j ↔ i < 0; (s) Used formulas: i = i0 * 2 + 1 - j