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

0 < j, 0 ≤ i2, j0 < j, i3 ≤ i2, 0 ≤ j0, 0 ≤ i3 ⊦ ∃ i. i2 = i3 + i

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: i2 - i3 abort: F discard: F computed: F

Additional information

Used simplifier rules: Sub-def: ⊦ i + (j - i) = j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i2 - i3