Theorem sbim 1234

Description: Implication inside and outside of substitution are equivalent.

Assertion

Ref	Expression
sbim	|- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))

Proof of Theorem sbim

Step	Hyp	Ref	Expression
1		sbi1 1232	. 2 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))
2		sbi2 1233	. 2 |- (([y / x]ph -> [y / x]ps) -> [y / x](ph -> ps))
3	1, 2	impbi 157	1 |- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))

Syntax hints:   -> wi 3   <-> wb 146  [wsbc 1170

This theorem is referenced by:  sbor 1235  sb19.21 1236  sban 1237  sbbi 1239  a4sbim 1244  sbequ8 1247  sbcimg 1970  tfinds2 3165

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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