Theorem sbba4 896

Description: Specialization of biconditional.

Assertion

Ref	Expression
sbba4	⊢ (∀x(φ ↔ ψ) → ([y / x]φ ↔ [y / x]ψ))

Proof of Theorem sbba4

Step	Hyp	Ref	Expression
1		stdpc4 869	. 2 ⊢ (∀x(φ ↔ ψ) → [y / x](φ ↔ ψ))
2		sbbi 890	. 2 ⊢ ([y / x](φ ↔ ψ) ↔ ([y / x]φ ↔ [y / x]ψ))
3	1, 2	sylib 173	1 ⊢ (∀x(φ ↔ ψ) → ([y / x]φ ↔ [y / x]ψ))

Syntax hints:   → wi 2   ↔ wb 127  ∀wal 672  [wsb 852

This theorem is referenced by:  bisbd 897  hbsb4t 906  sbco3 915

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

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