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Theorem sb5rf 1259
Description: Reversed substitution.
Hypothesis
Ref Expression
sb5rf.1 |- (ph -> A.yph)
Assertion
Ref Expression
sb5rf |- (ph <-> E.y(y = x /\ [y / x]ph))

Proof of Theorem sb5rf
StepHypRef Expression
1 sb5rf.1 . . . 4 |- (ph -> A.yph)
21sbid2 1253 . . 3 |- ([x / y][y / x]ph <-> ph)
3 sb1 1176 . . 3 |- ([x / y][y / x]ph -> E.y(y = x /\ [y / x]ph))
42, 3sylbir 201 . 2 |- (ph -> E.y(y = x /\ [y / x]ph))
5 sbequ12r 1182 . . . 4 |- (y = x -> ([y / x]ph <-> ph))
65biimpa 416 . . 3 |- ((y = x /\ [y / x]ph) -> ph)
71, 619.23ai 1064 . 2 |- (E.y(y = x /\ [y / x]ph) -> ph)
84, 7impbi 157 1 |- (ph <-> E.y(y = x /\ [y / x]ph))
Colors of variables: wff set class
Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 954   = wceq 956  E.wex 980  [wsbc 1170
This theorem is referenced by:  2sb5rf 1338
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-8 964  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-or 224  df-an 225  df-ex 981  df-sb 1172
Copyright terms: Public domain