Theorem sblbis 1240

Description: Introduce left biconditional inside of a substitution.

Hypothesis

Ref	Expression
sblbis.1	|- ([y / x]ph <-> ps)

Assertion

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

Proof of Theorem sblbis

Step	Hyp	Ref	Expression
1		sbbi 1239	. 2 |- ([y / x](ch <-> ph) <-> ([y / x]ch <-> [y / x]ph))
2		sblbis.1	. . 3 |- ([y / x]ph <-> ps)
3	2	bibi2i 608	. 2 |- (([y / x]ch <-> [y / x]ph) <-> ([y / x]ch <-> ps))
4	1, 3	bitr 173	1 |- ([y / x](ch <-> ph) <-> ([y / x]ch <-> ps))

Syntax hints:   <-> wb 146  [wsbc 1170

This theorem is referenced by:  sb8eu 1390

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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