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Proof step 11 from proof div2-nls (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = i0 * 2 - j + 1

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)%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) = j; (s)% ⊦ 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 < j + i0; (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ (* i2) * i < i + j0 ↔ (i2 - 1) * i < j0; (s)% ⊦ i < j → (j ≤ i ↔ false); (as) Used formulas: i = i0 * 2 - j + 1