Theorem sbrbif 893

Description: Introduce right biconditional inside of a substitution.

Hypotheses

Ref	Expression
sbrbif.1	⊢ (χ → ∀xχ)
sbrbif.2	⊢ ([y / x]φ ↔ ψ)

Assertion

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

Proof of Theorem sbrbif

Step	Hyp	Ref	Expression
1		sbrbif.2	. . 3 ⊢ ([y / x]φ ↔ ψ)
2	1	sbrbis 892	. 2 ⊢ ([y / x](φ ↔ χ) ↔ (ψ ↔ [y / x]χ))
3		sbrbif.1	. . . 4 ⊢ (χ → ∀xχ)
4	3	sbf 870	. . 3 ⊢ ([y / x]χ ↔ χ)
5	4	bibi2i 460	. 2 ⊢ ((ψ ↔ [y / x]χ) ↔ (ψ ↔ χ))
6	2, 5	bitr 151	1 ⊢ ([y / x](φ ↔ χ) ↔ (ψ ↔ χ))

Syntax hints:   → wi 2   ↔ 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

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

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