Theorem sbrbis 2073

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)

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 2071	. 2 ⊢ ([y / x](φ ↔ χ) ↔ ([y / x]φ ↔ [y / x]χ))
2		sbrbis.1	. . 3 ⊢ ([y / x]φ ↔ ψ)
3	2	bibi1i 305	. 2 ⊢ (([y / x]φ ↔ [y / x]χ) ↔ (ψ ↔ [y / x]χ))
4	1, 3	bitri 240	1 ⊢ ([y / x](φ ↔ χ) ↔ (ψ ↔ [y / x]χ))

Syntax hints:   ↔ wb 176  [wsb 1648

This theorem was proved from 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 theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  sbrbif  2074  sbabel  2516