Theorem sbrbis 892

Description: Introduce right biconditional inside of a substitution.

Hypothesis

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

Assertion

Ref	Expression
sbrbis	⊢ ([y / x](φ ↔ χ) ↔ (ψ ↔ [y / x]χ))

Proof of Theorem sbrbis

Step	Hyp	Ref	Expression
1		sbbi 890	. 2 ⊢ ([y / x](φ ↔ χ) ↔ ([y / x]φ ↔ [y / x]χ))
2		sbrbis.1	. . 3 ⊢ ([y / x]φ ↔ ψ)
3	2	bibi1i 461	. 2 ⊢ (([y / x]φ ↔ [y / x]χ) ↔ (ψ ↔ [y / x]χ))
4	1, 3	bitr 151	1 ⊢ ([y / x](φ ↔ χ) ↔ (ψ ↔ [y / x]χ))

Syntax hints:   ↔ wb 127  [wsb 852

This theorem is referenced by:  sbrbif 893

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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