Metamath Proof Explorer


Theorem rexcom4b

Description: Specialized existential commutation lemma. (Contributed by Jeff Madsen, 1-Jun-2011)

Ref Expression
Hypothesis rexcom4b.1 𝐵 ∈ V
Assertion rexcom4b ( ∃ 𝑥𝑦𝐴 ( 𝜑𝑥 = 𝐵 ) ↔ ∃ 𝑦𝐴 𝜑 )

Proof

Step Hyp Ref Expression
1 rexcom4b.1 𝐵 ∈ V
2 rexcom4a ( ∃ 𝑥𝑦𝐴 ( 𝜑𝑥 = 𝐵 ) ↔ ∃ 𝑦𝐴 ( 𝜑 ∧ ∃ 𝑥 𝑥 = 𝐵 ) )
3 1 isseti 𝑥 𝑥 = 𝐵
4 3 biantru ( 𝜑 ↔ ( 𝜑 ∧ ∃ 𝑥 𝑥 = 𝐵 ) )
5 4 rexbii ( ∃ 𝑦𝐴 𝜑 ↔ ∃ 𝑦𝐴 ( 𝜑 ∧ ∃ 𝑥 𝑥 = 𝐵 ) )
6 2 5 bitr4i ( ∃ 𝑥𝑦𝐴 ( 𝜑𝑥 = 𝐵 ) ↔ ∃ 𝑦𝐴 𝜑 )