Theorem bibi1i 461

Description: Inference adding a biconditional to the right in an equivalence.

Hypothesis

Ref	Expression
bibi.a	⊢ (φ ↔ ψ)

Assertion

Ref	Expression
bibi1i	⊢ ((φ ↔ χ) ↔ (ψ ↔ χ))

Proof of Theorem bibi1i

Step	Hyp	Ref	Expression
1		bicom 398	. 2 ⊢ ((φ ↔ χ) ↔ (χ ↔ φ))
2		bibi.a	. . 3 ⊢ (φ ↔ ψ)
3	2	bibi2i 460	. 2 ⊢ ((χ ↔ φ) ↔ (χ ↔ ψ))
4		bicom 398	. 2 ⊢ ((χ ↔ ψ) ↔ (ψ ↔ χ))
5	1, 3, 4	3bitr 155	1 ⊢ ((φ ↔ χ) ↔ (ψ ↔ χ))

Syntax hints:   ↔ wb 127

This theorem is referenced by:  bibi12i 462  sbrbis 892  axac 1085  sbabel 1189  aceq1 3552  aceq0 3553  zfcndac 3765

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-an 198

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