Theorem a4sbbi 1245

Description: Specialization of biconditional.

Assertion

Ref	Expression
a4sbbi	|- (A.x(ph <-> ps) -> ([y / x]ph <-> [y / x]ps))

Proof of Theorem a4sbbi

Step	Hyp	Ref	Expression
1		stdpc4 1185	. 2 |- (A.x(ph <-> ps) -> [y / x](ph <-> ps))
2		sbbi 1239	. 2 |- ([y / x](ph <-> ps) <-> ([y / x]ph <-> [y / x]ps))
3	1, 2	sylib 198	1 |- (A.x(ph <-> ps) -> ([y / x]ph <-> [y / x]ps))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 954  [wsbc 1170

This theorem is referenced by:  sbbid 1246  hbsb4t 1249  sbco3 1257

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-10 966  ax-12 968  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172

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