Theorem sbim 2065

Description: Implication inside and outside of substitution are equivalent. (Contributed by NM, 5-Aug-1993.)

Assertion

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

Proof of Theorem sbim

Step	Hyp	Ref	Expression
1		sbi1 2063	. 2 |- ([y / x](ph -> ps) -> ([y / x]ph -> [y / x]ps))
2		sbi2 2064	. 2 |- (([y / x]ph -> [y / x]ps) -> [y / x](ph -> ps))
3	1, 2	impbii 180	1 |- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))

Syntax hints:   -> wi 4   <-> wb 176  [wsb 1648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  sbor  2066  sbrim  2067  sblim  2068  sban  2069  sbbi  2071  sbequ8  2079  sbcimg  3087