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Take arbitrary A, B, C, f: A ->partial B, and g: B -> C. We prove (1) (f o g): A ->partial C. i.e., by definition of ->partial , that
We know (3) from the definition of o ; we still have to show (4). Take arbitrary x, y0, y1 and assume
(3) (f o g) subset A ×C; (4) forall x, y0, y1: (<x, y0> in (f o g) /\ <x, y1> in (f o g)) => y0=y1.
We have to show y0 = y1.
(5) <x, y0> in (f o g); (6) <x, y1> in (f o g).
From (5), (6), and the definition of o , we know y0 in C, y1 in C, and
By (7), we have some b0 in B such that <x, b0> in f /\ <b0, y0> in g; by (8), we have some b1 in B such that <x, b1> in f /\ <b1, y0> in g. ...
(7) exists b in B: <x, b> in f /\ <b, y0> in g; (8) exists b in B: <x, b> in f /\ <b, y1> in g.