Metamath Proof Explorer


Theorem spc2gv

Description: Specialization with two quantifiers, using implicit substitution. (Contributed by NM, 27-Apr-2004)

Ref Expression
Hypothesis spc2egv.1 ⊢ ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) → ( 𝜑 ↔ 𝜓 ) )
Assertion spc2gv ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( ∀ 𝑥 ∀ 𝑦 𝜑 → 𝜓 ) )

Proof

Step Hyp Ref Expression
1 spc2egv.1 ⊢ ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) → ( 𝜑 ↔ 𝜓 ) )
2 1 notbid ⊢ ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) → ( ¬ 𝜑 ↔ ¬ 𝜓 ) )
3 2 spc2egv ⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( ¬ 𝜓 → ∃ 𝑥 ∃ 𝑦 ¬ 𝜑 ) )
4 2nalexn ⊢ ( ¬ ∀ 𝑥 ∀ 𝑦 𝜑 ↔ ∃ 𝑥 ∃ 𝑦 ¬ 𝜑 )
5 3 4 imbitrrdi ⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( ¬ 𝜓 → ¬ ∀ 𝑥 ∀ 𝑦 𝜑 ) )
6 5 con4d ⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( ∀ 𝑥 ∀ 𝑦 𝜑 → 𝜓 ) )