Theorem ax10o 1499

Description: Show that ax-10o 1500 can be derived from ax-10 1308. An open problem is whether this theorem can be derived from ax-10 1308 and the others when ax-11 1309 is replaced with ax-11o 1588. See theorem ax10 1501 for the rederivation of ax-10 1308 from ax10o 1499.

This theorem should not be referenced in any proof. Instead, use ax-10o 1500 below so that uses of ax-10o 1500 can be more easily identified.

Assertion

Ref	Expression
ax10o	|- (A.x x = y -> (A.xph -> A.yph))

Proof of Theorem ax10o

Step	Hyp	Ref	Expression
1		ax-10 1308	. 2 |- (A.x x = y -> A.y y = x)
2		ax-11 1309	. . . 4 |- (y = x -> (A.xph -> A.y(y = x -> ph)))
3	2	equcoms 1489	. . 3 |- (x = y -> (A.xph -> A.y(y = x -> ph)))
4	3	a4s 1330	. 2 |- (A.x x = y -> (A.xph -> A.y(y = x -> ph)))
5		pm2.27 76	. . 3 |- (y = x -> ((y = x -> ph) -> ph))
6	5	al2imi 1341	. 2 |- (A.y y = x -> (A.y(y = x -> ph) -> A.yph))
7	1, 4, 6	sylsyld 32	1 |- (A.x x = y -> (A.xph -> A.yph))

Syntax hints:   -> wi 3  A.wal 1296   = wceq 1298

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 1305  ax-8 1306  ax-10 1308  ax-11 1309  ax-12 1310  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481

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