Metamath Proof Explorer


Theorem 2ralbiim

Description: Split a biconditional and distribute two restricted universal quantifiers, analogous to 2albiim and ralbiim . (Contributed by Alexander van der Vekens, 2-Jul-2017)

Ref Expression
Assertion 2ralbiim ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜑 ↔ 𝜓 ) ↔ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜑 → 𝜓 ) ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜓 → 𝜑 ) ) )

Proof

Step Hyp Ref Expression
1 ralbiim ⊢ ( ∀ 𝑦 ∈ 𝐵 ( 𝜑 ↔ 𝜓 ) ↔ ( ∀ 𝑦 ∈ 𝐵 ( 𝜑 → 𝜓 ) ∧ ∀ 𝑦 ∈ 𝐵 ( 𝜓 → 𝜑 ) ) )
2 1 ralbii ⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜑 ↔ 𝜓 ) ↔ ∀ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ( 𝜑 → 𝜓 ) ∧ ∀ 𝑦 ∈ 𝐵 ( 𝜓 → 𝜑 ) ) )
3 r19.26 ⊢ ( ∀ 𝑥 ∈ 𝐴 ( ∀ 𝑦 ∈ 𝐵 ( 𝜑 → 𝜓 ) ∧ ∀ 𝑦 ∈ 𝐵 ( 𝜓 → 𝜑 ) ) ↔ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜑 → 𝜓 ) ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜓 → 𝜑 ) ) )
4 2 3 bitri ⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜑 ↔ 𝜓 ) ↔ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜑 → 𝜓 ) ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ( 𝜓 → 𝜑 ) ) )