Theorem sb5 1252

Description: Equivalence for substitution. Similar to Theorem 6.1 of [Quine] p. 40.

Assertion

Ref	Expression
sb5	|- ([y / x]ph <-> E.x(x = y /\ ph))

Distinct variable group:   x,y

Proof of Theorem sb5

Step	Hyp	Ref	Expression
1		sb6 1251	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
2		sb56 1250	. 2 |- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))
3	1, 2	bitr4 176	1 |- ([y / x]ph <-> E.x(x = y /\ ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 950  E.wex 956   = wceq 1099  [wsbc 1153

This theorem is referenced by:  2sb5 1317  sb7 1322  sbelx 1326

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-gen 955  ax-9 1102  ax-16 1194  ax-11o 1202

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 957  df-sb 1155

Copyright terms: Public domain