Theorem bibi1i 305

Description: Inference adding a biconditional to the right in an equivalence. (Contributed by NM, 5-Aug-1993.)

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 191	. 2 ⊢ ((φ ↔ χ) ↔ (χ ↔ φ))
2		bibi.a	. . 3 ⊢ (φ ↔ ψ)
3	2	bibi2i 304	. 2 ⊢ ((χ ↔ φ) ↔ (χ ↔ ψ))
4		bicom 191	. 2 ⊢ ((χ ↔ ψ) ↔ (ψ ↔ χ))
5	1, 3, 4	3bitri 262	1 ⊢ ((φ ↔ χ) ↔ (ψ ↔ χ))

This proof depends on syntax axioms:   ↔ wb 176

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177

This theorem is used by:  bibi12i  306  biluk  899  xorass  1308  hadbi  1387  sbrbis  2073  ssequn1  3434  axssetprim  4093