Metamath Proof Explorer


Theorem sbcbiVD

Description: Implication form of sbcbii . The following User's Proof is a Virtual Deduction proof completed automatically by the tools program completeusersproof.cmd, which invokes Mel L. O'Cat's mmj2 and Norm Megill's Metamath Proof Assistant. sbcbi is sbcbiVD without virtual deductions and was automatically derived from sbcbiVD .

1:: |- (. A e. B ->. A e. B ).
2:: |- (. A e. B ,. A. x ( ph <-> ps ) ->. A. x ( ph <-> ps ) ).
3:1,2: |- (. A e. B ,. A. x ( ph <-> ps ) ->. [. A / x ]. ( ph <-> ps ) ).
4:1,3: |- (. A e. B ,. A. x ( ph <-> ps ) ->. ( [. A / x ]. ph <-> [. A / x ]. ps ) ).
5:4: |- (. A e. B ->. ( A. x ( ph <-> ps ) -> ( [. A / x ]. ph <-> [. A / x ]. ps ) ) ).
qed:5: |- ( A e. B -> ( A. x ( ph <-> ps ) -> ( [. A / x ]. ph <-> [. A / x ]. ps ) ) )
(Contributed by Alan Sare, 18-Mar-2012) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion sbcbiVD ( 𝐴𝐵 → ( ∀ 𝑥 ( 𝜑𝜓 ) → ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) ) )

Proof

Step Hyp Ref Expression
1 idn1 (    𝐴𝐵    ▶    𝐴𝐵    )
2 idn2 (    𝐴𝐵    ,   𝑥 ( 𝜑𝜓 )    ▶   𝑥 ( 𝜑𝜓 )    )
3 spsbc ( 𝐴𝐵 → ( ∀ 𝑥 ( 𝜑𝜓 ) → [ 𝐴 / 𝑥 ] ( 𝜑𝜓 ) ) )
4 1 2 3 e12 (    𝐴𝐵    ,   𝑥 ( 𝜑𝜓 )    ▶    [ 𝐴 / 𝑥 ] ( 𝜑𝜓 )    )
5 sbcbig ( 𝐴𝐵 → ( [ 𝐴 / 𝑥 ] ( 𝜑𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) ) )
6 5 biimpd ( 𝐴𝐵 → ( [ 𝐴 / 𝑥 ] ( 𝜑𝜓 ) → ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) ) )
7 1 4 6 e12 (    𝐴𝐵    ,   𝑥 ( 𝜑𝜓 )    ▶    ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 )    )
8 7 in2 (    𝐴𝐵    ▶    ( ∀ 𝑥 ( 𝜑𝜓 ) → ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) )    )
9 8 in1 ( 𝐴𝐵 → ( ∀ 𝑥 ( 𝜑𝜓 ) → ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) ) )