Metamath Proof Explorer


Theorem csbie2df

Description: Conversion of implicit substitution to explicit class substitution. This version of csbiedf avoids a disjointness condition on x , A and x , D by substituting twice. Deduction form of csbie2 . (Contributed by AV, 29-Mar-2024)

Ref Expression
Hypotheses csbie2df.p x φ
csbie2df.c φ _ x C
csbie2df.d φ _ x D
csbie2df.a φ A V
csbie2df.1 φ x = y B = C
csbie2df.2 φ y = A C = D
Assertion csbie2df φ A / x B = D

Proof

Step Hyp Ref Expression
1 csbie2df.p x φ
2 csbie2df.c φ _ x C
3 csbie2df.d φ _ x D
4 csbie2df.a φ A V
5 csbie2df.1 φ x = y B = C
6 csbie2df.2 φ y = A C = D
7 eqidd φ D = D
8 dfsbcq y = A [˙y / x]˙ B = D [˙A / x]˙ B = D
9 sbceqg A V [˙A / x]˙ B = D A / x B = A / x D
10 9 adantr A V φ [˙A / x]˙ B = D A / x B = A / x D
11 csbtt A V _ x D A / x D = D
12 3 11 sylan2 A V φ A / x D = D
13 12 eqeq2d A V φ A / x B = A / x D A / x B = D
14 10 13 bitrd A V φ [˙A / x]˙ B = D A / x B = D
15 4 14 mpancom φ [˙A / x]˙ B = D A / x B = D
16 8 15 sylan9bb y = A φ [˙y / x]˙ B = D A / x B = D
17 16 pm5.74da y = A φ [˙y / x]˙ B = D φ A / x B = D
18 6 eqeq1d φ y = A C = D D = D
19 18 expcom y = A φ C = D D = D
20 19 pm5.74d y = A φ C = D φ D = D
21 sbsbc y x B = D [˙y / x]˙ B = D
22 2 3 nfeqd φ x C = D
23 5 eqeq1d φ x = y B = D C = D
24 23 ex φ x = y B = D C = D
25 1 22 24 sbiedw φ y x B = D C = D
26 21 25 bitr3id φ [˙y / x]˙ B = D C = D
27 26 pm5.74i φ [˙y / x]˙ B = D φ C = D
28 17 20 27 vtoclbg A V φ A / x B = D φ D = D
29 7 28 mpbiri A V φ A / x B = D
30 4 29 mpcom φ A / x B = D