Theorem sb56 1806

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 1686. (Contributed by NM, 14-Apr-2008.)

Assertion

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

Distinct variable group:   x, y

Allowed substitution hints:   ph( x, y)

Proof of Theorem sb56

Step	Hyp	Ref	Expression
1		hba1 1473	. 2 |- ( A. x ( x = y -> ph ) -> A. x A. x ( x = y -> ph ) )
2		ax11v 1748	. . 3 |- ( x = y -> ( ph -> A. x ( x = y -> ph ) ) )
3		ax-4 1440	. . . 4 |- ( A. x ( x = y -> ph ) -> ( x = y -> ph ) )
4	3	com12 30	. . 3 |- ( x = y -> ( A. x ( x = y -> ph ) -> ph ) )
5	2, 4	impbid 127	. 2 |- ( x = y -> ( ph <-> A. x ( x = y -> ph ) ) )
6	1, 5	equsex 1656	1 |- ( E. x ( x = y /\ ph ) <-> A. x ( x = y -> ph ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   A.wal 1282   E.wex 1421

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  sb6  1807  sb5  1808  alexeq  2721