Theorem sbalv 999

Description: Quantify with new variable inside substitution.

Hypothesis

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

Assertion

Ref	Expression
sbalv	|- ([y / x]A.zph <-> A.zps)

Distinct variable group(s):   x,z   y,z

Proof of Theorem sbalv

Step	Hyp	Ref	Expression
1		sbal 997	. 2 |- ([y / x]A.zph <-> A.z[y / x]ph)
2		sbalv.1	. . 3 |- ([y / x]ph <-> ps)
3	2	bial 695	. 2 |- (A.z[y / x]ph <-> A.zps)
4	1, 3	bitr 151	1 |- ([y / x]A.zph <-> A.zps)

Syntax hints:   <-> wb 127  A.wal 672  [wsb 852

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  ax-16 922

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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