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Proof step 9 from proof modeven-neg (previous step)

1 + i0 * 2 = i1 * 2 + j0, abs(j0) < 2, i0 < 0, j0 ≤ 0 ⊦ j0 = ~1, j0 = ~1

User applied rule cut formula

And closed the goal!

Rule argument

fma: j0 = 0

Additional information

Used simplifier rules: lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%minus-one: ⊦ i < 0 → (~2 < i ↔ i = ~1); from specification int-basic1%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%odd-even: ⊦ 2 * i ≠ 2 * j + 1; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used formulas: j0 = 0