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Proof step 10 from proof div-abs-01 (previous step)

j + i * i0 = 0, j < i * i0, j < 0, 0 < i * i0, j + 1 < i * i0 ⊦ ∃ i0. i * i0 = j

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: ~ i0 abort: F discard: T computed: F

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: ~ i0