Theorem sbim 886

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 884	. 2 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))
2		sbi2 885	. 2 |- (([y / x]ph -> [y / x]ps) -> [y / x](ph -> ps))
3	1, 2	impbi 139	1 |- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))

Syntax hints:   -> wi 2   <-> wb 127  [wsb 852

This theorem is referenced by:  sbor 887  sb19.21 888  sban 889  sbbi 890  sbia4 895  sbequ8 902  sbcim 1460  tfinds2 2405

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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