Theorem sbor 887

Description: Logical OR inside and outside of substitution are equivalent.

Assertion

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

Proof of Theorem sbor

Step	Hyp	Ref	Expression
1		sbim 886	. . 3 ⊢ ([y / x](¬ φ → ψ) ↔ ([y / x] ¬ φ → [y / x]ψ))
2		sbn 882	. . . 4 ⊢ ([y / x] ¬ φ ↔ ¬ [y / x]φ)
3	2	imbi1i 161	. . 3 ⊢ (([y / x] ¬ φ → [y / x]ψ) ↔ (¬ [y / x]φ → [y / x]ψ))
4	1, 3	bitr 151	. 2 ⊢ ([y / x](¬ φ → ψ) ↔ (¬ [y / x]φ → [y / x]ψ))
5		df-or 197	. . 3 ⊢ ((φ ∨ ψ) ↔ (¬ φ → ψ))
6	5	bisb 855	. 2 ⊢ ([y / x](φ ∨ ψ) ↔ [y / x](¬ φ → ψ))
7		df-or 197	. 2 ⊢ (([y / x]φ ∨ [y / x]ψ) ↔ (¬ [y / x]φ → [y / x]ψ))
8	4, 6, 7	3bitr4 158	1 ⊢ ([y / x](φ ∨ ψ) ↔ ([y / x]φ ∨ [y / x]ψ))

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∨ wo 195  [wsb 852

This theorem is referenced by:  sbcor 1462  unab 1691

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-or 197  df-an 198  df-ex 679  df-sb 853

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