Theorem sbalv 999

Description: Quantify with new variable inside substitution.

Hypothesis

Ref	Expression
sbalv.1	⊢ ([y / x]φ ↔ ψ)

Assertion

Ref	Expression
sbalv	⊢ ([y / x]∀zφ ↔ ∀zψ)

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

Proof of Theorem sbalv

Step	Hyp	Ref	Expression
1		sbal 997	. 2 ⊢ ([y / x]∀zφ ↔ ∀z[y / x]φ)
2		sbalv.1	. . 3 ⊢ ([y / x]φ ↔ ψ)
3	2	bial 695	. 2 ⊢ (∀z[y / x]φ ↔ ∀zψ)
4	1, 3	bitr 151	1 ⊢ ([y / x]∀zφ ↔ ∀zψ)

Syntax hints:   ↔ wb 127  ∀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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