Theorem sbrbif 2074

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypotheses

Ref	Expression
sbrbif.1	⊢ Ⅎxχ
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 2073	. 2 ⊢ ([y / x](φ ↔ χ) ↔ (ψ ↔ [y / x]χ))
3		sbrbif.1	. . . 4 ⊢ Ⅎxχ
4	3	sbf 2026	. . 3 ⊢ ([y / x]χ ↔ χ)
5	4	bibi2i 304	. 2 ⊢ ((ψ ↔ [y / x]χ) ↔ (ψ ↔ χ))
6	2, 5	bitri 240	1 ⊢ ([y / x](φ ↔ χ) ↔ (ψ ↔ χ))

This proof depends on syntax axioms:   ↔ wb 176  Ⅎwnf 1544  [wsb 1648

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is used by: (None)