Theorem ax5o 1015

Description: Show that the original axiom ax-5o 1016 can be derived from ax-5 1001 and others. See ax5 1017 for the rederivation of ax-5 1001 from ax-5o 1016.

Part of the proof is based on the proof of Lemma 22 of [Monk2] p. 114.

This theorem should not be referenced in any proof. Instead, use ax-5o 1016 below so that uses of ax-5o 1016 can be more easily identified.

Assertion

Ref	Expression
ax5o	|- (A.x(A.xph -> ps) -> (A.xph -> A.xps))

Proof of Theorem ax5o

Step	Hyp	Ref	Expression
1		ax-5 1001	. 2 |- (A.x(A.xph -> ps) -> (A.xA.xph -> A.xps))
2		ax-4 1014	. . . 4 |- (A.x -. A.xph -> -. A.xph)
3	2	con2i 102	. . 3 |- (A.xph -> -. A.x -. A.xph)
4		ax-6 1002	. . 3 |- (-. A.x -. A.xph -> A.x -. A.x -. A.xph)
5		ax-6 1002	. . . . . 6 |- (-. A.xph -> A.x -. A.xph)
6	5	con1i 100	. . . . 5 |- (-. A.x -. A.xph -> A.xph)
7	6	ax-gen 1004	. . . 4 |- A.x(-. A.x -. A.xph -> A.xph)
8		ax-5 1001	. . . 4 |- (A.x(-. A.x -. A.xph -> A.xph) -> (A.x -. A.x -. A.xph -> A.xA.xph))
9	7, 8	ax-mp 7	. . 3 |- (A.x -. A.x -. A.xph -> A.xA.xph)
10	3, 4, 9	3syl 20	. 2 |- (A.xph -> A.xA.xph)
11	1, 10	syl5 21	1 |- (A.x(A.xph -> ps) -> (A.xph -> A.xps))

Syntax hints:  -. wn 2   -> wi 3  A.wal 995

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-5 1001  ax-6 1002  ax-gen 1004  ax-4 1014

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