Theorem sbbid 2078

Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 5-Aug-1993.)

Hypotheses

Ref	Expression
sbbid.1	⊢ Ⅎxφ
sbbid.2	⊢ (φ → (ψ ↔ χ))

Assertion

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

Proof of Theorem sbbid

Step	Hyp	Ref	Expression
1		sbbid.1	. . 3 ⊢ Ⅎxφ
2		sbbid.2	. . 3 ⊢ (φ → (ψ ↔ χ))
3	1, 2	alrimi 1765	. 2 ⊢ (φ → ∀x(ψ ↔ χ))
4		spsbbi 2077	. 2 ⊢ (∀x(ψ ↔ χ) → ([y / x]ψ ↔ [y / x]χ))
5	3, 4	syl 15	1 ⊢ (φ → ([y / x]ψ ↔ [y / x]χ))

Syntax hints:   → wi 4   ↔ wb 176  ∀wal 1540  Ⅎwnf 1544  [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:  sbcom  2089  sbcom2  2114