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**OReilly** These are the axioms of incidence from the book that I have:

I.1: Every line contains at least two distinct points.

I.2: There exists at least one line that contains two points.

I.3: There exists at most one line that containts two distinct points.

I.4: Every plane contains at least three non-colinear points.

I.5: There exists at least one plane that contains three points.

I.6: There exists at most one plane that contains three non-colinear points.

I.7: If two distinct points of some line belongs to one plane, then every point of that line belongs to same plane.

I.8: If two distinct planes have one common point, then they have at least one more common point.

I.9: There are four non-coplanar points.

I have to prove this:

"If we substitute axiom I.4 by axiom I.4': 'Every plane contains at least one point', then prove that consequence from given group of axioms (which I have typed) is axiom I.4"

My proof goes like this:

From I.9 we have three non-colinear points A,B,C which forms one plane $\displaystyle \alpha $ by axiom I.6 and one point D not contained by plane $\displaystyle \alpha $.

Any two points of plane $\displaystyle \alpha $ can form with point D another plane $\displaystyle \beta $.

Any plane different then $\displaystyle \alpha $ and $\displaystyle \beta $ must intersect one of them. By axiom I.4' that plane contains at least one point and by intersecting one of two planes $\displaystyle \alpha $ or $\displaystyle \beta $ it has two more distinct points by axiom I.8 which means that every plane has three non-colinear points.

Is my proof correct?