Theorem sbbi 890

Description: Equivalence inside and outside of a substitution are equivalent.

Assertion

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

Proof of Theorem sbbi

Step	Hyp	Ref	Expression
1		bi 396	. . 3 |- ((ph <-> ps) <-> ((ph -> ps) /\ (ps -> ph)))
2	1	bisb 855	. 2 |- ([y / x](ph <-> ps) <-> [y / x]((ph -> ps) /\ (ps -> ph)))
3		sbim 886	. . . 4 |- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))
4		sbim 886	. . . 4 |- ([y / x](ps -> ph) <-> ([y / x]ps -> [y / x]ph))
5	3, 4	anbi12i 369	. . 3 |- (([y / x](ph -> ps) /\ [y / x](ps -> ph)) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
6		sban 889	. . 3 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x](ph -> ps) /\ [y / x](ps -> ph)))
7		bi 396	. . 3 |- (([y / x]ph <-> [y / x]ps) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
8	5, 6, 7	3bitr4 158	. 2 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x]ph <-> [y / x]ps))
9	2, 8	bitr 151	1 |- ([y / x](ph <-> ps) <-> ([y / x]ph <-> [y / x]ps))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196  [wsb 852

This theorem is referenced by:  sblbis 891  sbrbis 892  sbba4 896  sbco 910  sbal 997  sbabel 1189  sbcbi 1463

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

metamath.org