Theorem bisbd 897

Description: Deduction substituting both sides of a biconditional.

Hypotheses

Ref	Expression
bisbd.1	⊢ (φ → ∀xφ)
bisbd.2	⊢ (φ → (ψ ↔ χ))

Assertion

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

Proof of Theorem bisbd

Step	Hyp	Ref	Expression
1		bisbd.1	. . 3 ⊢ (φ → ∀xφ)
2		bisbd.2	. . 3 ⊢ (φ → (ψ ↔ χ))
3	1, 2	19.21ai 740	. 2 ⊢ (φ → ∀x(ψ ↔ χ))
4		sbba4 896	. 2 ⊢ (∀x(ψ ↔ χ) → ([y / x]ψ ↔ [y / x]χ))
5	3, 4	syl 12	1 ⊢ (φ → ([y / x]ψ ↔ [y / x]χ))

Syntax hints:   → wi 2   ↔ wb 127  ∀wal 672  [wsb 852

This theorem is referenced by:  sbcom 916  sbcom2 992

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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