Metamath Proof Explorer


Theorem ceqsex2

Description: Elimination of two existential quantifiers, using implicit substitution. (Contributed by Scott Fenton, 7-Jun-2006)

Ref Expression
Hypotheses ceqsex2.1 𝑥 𝜓
ceqsex2.2 𝑦 𝜒
ceqsex2.3 𝐴 ∈ V
ceqsex2.4 𝐵 ∈ V
ceqsex2.5 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
ceqsex2.6 ( 𝑦 = 𝐵 → ( 𝜓𝜒 ) )
Assertion ceqsex2 ( ∃ 𝑥𝑦 ( 𝑥 = 𝐴𝑦 = 𝐵𝜑 ) ↔ 𝜒 )

Proof

Step Hyp Ref Expression
1 ceqsex2.1 𝑥 𝜓
2 ceqsex2.2 𝑦 𝜒
3 ceqsex2.3 𝐴 ∈ V
4 ceqsex2.4 𝐵 ∈ V
5 ceqsex2.5 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
6 ceqsex2.6 ( 𝑦 = 𝐵 → ( 𝜓𝜒 ) )
7 3anass ( ( 𝑥 = 𝐴𝑦 = 𝐵𝜑 ) ↔ ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵𝜑 ) ) )
8 7 exbii ( ∃ 𝑦 ( 𝑥 = 𝐴𝑦 = 𝐵𝜑 ) ↔ ∃ 𝑦 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵𝜑 ) ) )
9 19.42v ( ∃ 𝑦 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵𝜑 ) ) ↔ ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ( 𝑦 = 𝐵𝜑 ) ) )
10 8 9 bitri ( ∃ 𝑦 ( 𝑥 = 𝐴𝑦 = 𝐵𝜑 ) ↔ ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ( 𝑦 = 𝐵𝜑 ) ) )
11 10 exbii ( ∃ 𝑥𝑦 ( 𝑥 = 𝐴𝑦 = 𝐵𝜑 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ( 𝑦 = 𝐵𝜑 ) ) )
12 nfv 𝑥 𝑦 = 𝐵
13 12 1 nfan 𝑥 ( 𝑦 = 𝐵𝜓 )
14 13 nfex 𝑥𝑦 ( 𝑦 = 𝐵𝜓 )
15 5 anbi2d ( 𝑥 = 𝐴 → ( ( 𝑦 = 𝐵𝜑 ) ↔ ( 𝑦 = 𝐵𝜓 ) ) )
16 15 exbidv ( 𝑥 = 𝐴 → ( ∃ 𝑦 ( 𝑦 = 𝐵𝜑 ) ↔ ∃ 𝑦 ( 𝑦 = 𝐵𝜓 ) ) )
17 14 3 16 ceqsex ( ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ( 𝑦 = 𝐵𝜑 ) ) ↔ ∃ 𝑦 ( 𝑦 = 𝐵𝜓 ) )
18 2 4 6 ceqsex ( ∃ 𝑦 ( 𝑦 = 𝐵𝜓 ) ↔ 𝜒 )
19 11 17 18 3bitri ( ∃ 𝑥𝑦 ( 𝑥 = 𝐴𝑦 = 𝐵𝜑 ) ↔ 𝜒 )