Metamath Proof Explorer


Theorem eqsbc2VD

Description: Virtual deduction proof of eqsbc2 . (Contributed by Alan Sare, 24-Oct-2011) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion eqsbc2VD ( 𝐴𝐵 → ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥𝐶 = 𝐴 ) )

Proof

Step Hyp Ref Expression
1 idn1 (    𝐴𝐵    ▶    𝐴𝐵    )
2 eqsbc1 ( 𝐴𝐵 → ( [ 𝐴 / 𝑥 ] 𝑥 = 𝐶𝐴 = 𝐶 ) )
3 1 2 e1a (    𝐴𝐵    ▶    ( [ 𝐴 / 𝑥 ] 𝑥 = 𝐶𝐴 = 𝐶 )    )
4 eqcom ( 𝐶 = 𝑥𝑥 = 𝐶 )
5 4 sbcbii ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥[ 𝐴 / 𝑥 ] 𝑥 = 𝐶 )
6 5 a1i ( 𝐴𝐵 → ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥[ 𝐴 / 𝑥 ] 𝑥 = 𝐶 ) )
7 1 6 e1a (    𝐴𝐵    ▶    ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥[ 𝐴 / 𝑥 ] 𝑥 = 𝐶 )    )
8 idn2 (    𝐴𝐵    ,    [ 𝐴 / 𝑥 ] 𝐶 = 𝑥    ▶    [ 𝐴 / 𝑥 ] 𝐶 = 𝑥    )
9 biimp ( ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥[ 𝐴 / 𝑥 ] 𝑥 = 𝐶 ) → ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥[ 𝐴 / 𝑥 ] 𝑥 = 𝐶 ) )
10 7 8 9 e12 (    𝐴𝐵    ,    [ 𝐴 / 𝑥 ] 𝐶 = 𝑥    ▶    [ 𝐴 / 𝑥 ] 𝑥 = 𝐶    )
11 biimp ( ( [ 𝐴 / 𝑥 ] 𝑥 = 𝐶𝐴 = 𝐶 ) → ( [ 𝐴 / 𝑥 ] 𝑥 = 𝐶𝐴 = 𝐶 ) )
12 3 10 11 e12 (    𝐴𝐵    ,    [ 𝐴 / 𝑥 ] 𝐶 = 𝑥    ▶    𝐴 = 𝐶    )
13 eqcom ( 𝐴 = 𝐶𝐶 = 𝐴 )
14 12 13 e2bi (    𝐴𝐵    ,    [ 𝐴 / 𝑥 ] 𝐶 = 𝑥    ▶    𝐶 = 𝐴    )
15 14 in2 (    𝐴𝐵    ▶    ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥𝐶 = 𝐴 )    )
16 idn2 (    𝐴𝐵    ,    𝐶 = 𝐴    ▶    𝐶 = 𝐴    )
17 16 13 e2bir (    𝐴𝐵    ,    𝐶 = 𝐴    ▶    𝐴 = 𝐶    )
18 biimpr ( ( [ 𝐴 / 𝑥 ] 𝑥 = 𝐶𝐴 = 𝐶 ) → ( 𝐴 = 𝐶[ 𝐴 / 𝑥 ] 𝑥 = 𝐶 ) )
19 3 17 18 e12 (    𝐴𝐵    ,    𝐶 = 𝐴    ▶    [ 𝐴 / 𝑥 ] 𝑥 = 𝐶    )
20 biimpr ( ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥[ 𝐴 / 𝑥 ] 𝑥 = 𝐶 ) → ( [ 𝐴 / 𝑥 ] 𝑥 = 𝐶[ 𝐴 / 𝑥 ] 𝐶 = 𝑥 ) )
21 7 19 20 e12 (    𝐴𝐵    ,    𝐶 = 𝐴    ▶    [ 𝐴 / 𝑥 ] 𝐶 = 𝑥    )
22 21 in2 (    𝐴𝐵    ▶    ( 𝐶 = 𝐴[ 𝐴 / 𝑥 ] 𝐶 = 𝑥 )    )
23 impbi ( ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥𝐶 = 𝐴 ) → ( ( 𝐶 = 𝐴[ 𝐴 / 𝑥 ] 𝐶 = 𝑥 ) → ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥𝐶 = 𝐴 ) ) )
24 15 22 23 e11 (    𝐴𝐵    ▶    ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥𝐶 = 𝐴 )    )
25 24 in1 ( 𝐴𝐵 → ( [ 𝐴 / 𝑥 ] 𝐶 = 𝑥𝐶 = 𝐴 ) )