Theorem sb56 1261

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 1168.

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 1000	. 2 |- (A.x(x = y -> ph) -> A.xA.x(x = y -> ph))
2		ax11v 1260	. . 3 |- (x = y -> (ph -> A.x(x = y -> ph)))
3		ax-4 970	. . . 4 |- (A.x(x = y -> ph) -> (x = y -> ph))
4	3	com12 11	. . 3 |- (x = y -> (A.x(x = y -> ph) -> ph))
5	2, 4	impbid 514	. 2 |- (x = y -> (ph <-> A.x(x = y -> ph)))
6	1, 5	equsex 1148	1 |- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 951   = wceq 953  E.wex 977

This theorem is referenced by:  sb6 1262  sb5 1263  alexeq 1876

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 960  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-16 1206  ax-11o 1213

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978

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