Theorem sb56 1643

Description: Two equivalent ways of expressing the proper substitution of y for x in ph, when x and y are distinct. Theorem 6.2 of [Quine] p. 40. The proof does not involve df-sb 1536.

Assertion

Ref	Expression
sb56	|- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))

Distinct variable group:   x,y

Proof of Theorem sb56

Step	Hyp	Ref	Expression
1		hba1 1350	. 2 |- (A.x(x = y -> ph) -> A.xA.x(x = y -> ph))
2		ax11v 1642	. . 3 |- (x = y -> (ph -> A.x(x = y -> ph)))
3		ax-4 1319	. . . 4 |- (A.x(x = y -> ph) -> (x = y -> ph))
4	3	com12 14	. . 3 |- (x = y -> (A.x(x = y -> ph) -> ph))
5	2, 4	impbid 574	. 2 |- (x = y -> (ph <-> A.x(x = y -> ph)))
6	1, 5	equsex 1513	1 |- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))

Syntax hints:   -> wi 3   <-> wb 163   /\ wa 240  A.wal 1296   = wceq 1298  E.wex 1326

This theorem is referenced by:  sb6 1644  sb5 1645  alexeq 2390  pm13.196a 16378

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 1305  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-16 1580  ax-11o 1588

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327

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