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Metamath Proof Explorer

Theorem rexeqbidvv

Description: Version of rexeqbidv with additional disjoint variable conditions, not requiring ax-8 nor df-clel . (Contributed by Wolf Lammen, 25-Sep-2024)

	Ref	Expression
Hypotheses	raleqbidvv.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
	raleqbidvv.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion	rexeqbidvv	⊢ ( 𝜑 → ( ∃ 𝑥 ∈ 𝐴 𝜓 ↔ ∃ 𝑥 ∈ 𝐵 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		raleqbidvv.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		raleqbidvv.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
3	2	adantr	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
4	1 3	rexeqbidva	⊢ ( 𝜑 → ( ∃ 𝑥 ∈ 𝐴 𝜓 ↔ ∃ 𝑥 ∈ 𝐵 𝜒 ) )