Theorem sbim 1236

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 1234	. 2 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))
2		sbi2 1235	. 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 1172

This theorem is referenced by:  sbor 1237  sb19.21 1238  sban 1239  sbbi 1241  a4sbim 1246  sbequ8 1249  sbcimg 1973  tfinds2 3171

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-10 968  ax-12 970  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-11o 1220

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174

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