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Metamath Proof Explorer

Theorem r2exlem

Description: Lemma factoring out common proof steps in r2exf an r2ex . Introduced to reduce dependencies on axioms. (Contributed by Wolf Lammen, 10-Jan-2020)

		Ref	Expression
	Hypothesis	r2exlem.1	⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝜑 ) )
	Assertion	r2exlem	⊢ ( ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑥 ∃ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		r2exlem.1	⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝜑 ) )
2		exnal	⊢ ( ∃ 𝑥 ¬ ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝜑 ) ↔ ¬ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝜑 ) )
3	2 1	xchbinxr	⊢ ( ∃ 𝑥 ¬ ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝜑 ) ↔ ¬ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ¬ 𝜑 )
4		exnalimn	⊢ ( ∃ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝜑 ) ↔ ¬ ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝜑 ) )
5	4	exbii	⊢ ( ∃ 𝑥 ∃ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝜑 ) ↔ ∃ 𝑥 ¬ ∀ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝜑 ) )
6		ralnex2	⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ¬ ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 )
7	6	con2bii	⊢ ( ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ¬ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ¬ 𝜑 )
8	3 5 7	3bitr4ri	⊢ ( ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑥 ∃ 𝑦 ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝜑 ) )