Metamath Proof Explorer


Theorem csbhypf

Description: Introduce an explicit substitution into an implicit substitution hypothesis. See sbhypf for class substitution version. (Contributed by NM, 19-Dec-2008)

Ref Expression
Hypotheses csbhypf.1 𝑥 𝐴
csbhypf.2 𝑥 𝐶
csbhypf.3 ( 𝑥 = 𝐴𝐵 = 𝐶 )
Assertion csbhypf ( 𝑦 = 𝐴 𝑦 / 𝑥 𝐵 = 𝐶 )

Proof

Step Hyp Ref Expression
1 csbhypf.1 𝑥 𝐴
2 csbhypf.2 𝑥 𝐶
3 csbhypf.3 ( 𝑥 = 𝐴𝐵 = 𝐶 )
4 1 nfeq2 𝑥 𝑦 = 𝐴
5 nfcsb1v 𝑥 𝑦 / 𝑥 𝐵
6 5 2 nfeq 𝑥 𝑦 / 𝑥 𝐵 = 𝐶
7 4 6 nfim 𝑥 ( 𝑦 = 𝐴 𝑦 / 𝑥 𝐵 = 𝐶 )
8 eqeq1 ( 𝑥 = 𝑦 → ( 𝑥 = 𝐴𝑦 = 𝐴 ) )
9 csbeq1a ( 𝑥 = 𝑦𝐵 = 𝑦 / 𝑥 𝐵 )
10 9 eqeq1d ( 𝑥 = 𝑦 → ( 𝐵 = 𝐶 𝑦 / 𝑥 𝐵 = 𝐶 ) )
11 8 10 imbi12d ( 𝑥 = 𝑦 → ( ( 𝑥 = 𝐴𝐵 = 𝐶 ) ↔ ( 𝑦 = 𝐴 𝑦 / 𝑥 𝐵 = 𝐶 ) ) )
12 7 11 3 chvarfv ( 𝑦 = 𝐴 𝑦 / 𝑥 𝐵 = 𝐶 )