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Proof step 10 from proof div2-nle-02 (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = i0 * 2 - j

Additional information

Used simplifier rules: nummult-eq-01: j ≠ 0, j0 % j = 0 ⊦ i * (* j) = i0 * (* j0) ↔ i = i0 * j0 / j;% ⊦ 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% ⊦ 1 * i = i; (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 + i0 ≤ j; (s)% ⊦ i + 0 = i; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i + i = 2 * i; (s)% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); (s)% ⊦ i ≤ j → ¬ j < i; (s) Used formulas: i = i0 * 2 - j