Metamath Proof Explorer


Theorem rexcom4a

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

Ref Expression
Assertion rexcom4a ( ∃ 𝑥 ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ 𝜓 ) ↔ ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ ∃ 𝑥 𝜓 ) )

Proof

Step Hyp Ref Expression
1 rexcom4 ⊢ ( ∃ 𝑦 ∈ 𝐴 ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ↔ ∃ 𝑥 ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ 𝜓 ) )
2 19.42v ⊢ ( ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ↔ ( 𝜑 ∧ ∃ 𝑥 𝜓 ) )
3 2 rexbii ⊢ ( ∃ 𝑦 ∈ 𝐴 ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ↔ ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ ∃ 𝑥 𝜓 ) )
4 1 3 bitr3i ⊢ ( ∃ 𝑥 ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ 𝜓 ) ↔ ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ ∃ 𝑥 𝜓 ) )