Theorem sbbi 890

Description: Equivalence inside and outside of a substitution are equivalent.

Assertion

Ref	Expression
sbbi	⊢ ([y / x](φ ↔ ψ) ↔ ([y / x]φ ↔ [y / x]ψ))

Proof of Theorem sbbi

Step	Hyp	Ref	Expression
1		bi 396	. . 3 ⊢ ((φ ↔ ψ) ↔ ((φ → ψ) ∧ (ψ → φ)))
2	1	bisb 855	. 2 ⊢ ([y / x](φ ↔ ψ) ↔ [y / x]((φ → ψ) ∧ (ψ → φ)))
3		sbim 886	. . . 4 ⊢ ([y / x](φ → ψ) ↔ ([y / x]φ → [y / x]ψ))
4		sbim 886	. . . 4 ⊢ ([y / x](ψ → φ) ↔ ([y / x]ψ → [y / x]φ))
5	3, 4	anbi12i 369	. . 3 ⊢ (([y / x](φ → ψ) ∧ [y / x](ψ → φ)) ↔ (([y / x]φ → [y / x]ψ) ∧ ([y / x]ψ → [y / x]φ)))
6		sban 889	. . 3 ⊢ ([y / x]((φ → ψ) ∧ (ψ → φ)) ↔ ([y / x](φ → ψ) ∧ [y / x](ψ → φ)))
7		bi 396	. . 3 ⊢ (([y / x]φ ↔ [y / x]ψ) ↔ (([y / x]φ → [y / x]ψ) ∧ ([y / x]ψ → [y / x]φ)))
8	5, 6, 7	3bitr4 158	. 2 ⊢ ([y / x]((φ → ψ) ∧ (ψ → φ)) ↔ ([y / x]φ ↔ [y / x]ψ))
9	2, 8	bitr 151	1 ⊢ ([y / x](φ ↔ ψ) ↔ ([y / x]φ ↔ [y / x]ψ))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196  [wsb 852

This theorem is referenced by:  sblbis 891  sbrbis 892  sbba4 896  sbco 910  sbal 997  sbabel 1189  sbcbi 1463

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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