Metamath Proof Explorer


Theorem ssrexf

Description: Restricted existential quantification follows from a subclass relationship. (Contributed by Glauco Siliprandi, 20-Apr-2017)

Ref Expression
Hypotheses ssrexf.1 ⊢ Ⅎ 𝑥 𝐴
ssrexf.2 ⊢ Ⅎ 𝑥 𝐵
Assertion ssrexf ( 𝐴 ⊆ 𝐵 → ( ∃ 𝑥 ∈ 𝐴 𝜑 → ∃ 𝑥 ∈ 𝐵 𝜑 ) )

Proof

Step Hyp Ref Expression
1 ssrexf.1 ⊢ Ⅎ 𝑥 𝐴
2 ssrexf.2 ⊢ Ⅎ 𝑥 𝐵
3 1 2 nfss ⊢ Ⅎ 𝑥 𝐴 ⊆ 𝐵
4 ssel ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
5 4 anim1d ⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
6 3 5 eximd ⊢ ( 𝐴 ⊆ 𝐵 → ( ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ∃ 𝑥 ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
7 df-rex ⊢ ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) )
8 df-rex ⊢ ( ∃ 𝑥 ∈ 𝐵 𝜑 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) )
9 6 7 8 3imtr4g ⊢ ( 𝐴 ⊆ 𝐵 → ( ∃ 𝑥 ∈ 𝐴 𝜑 → ∃ 𝑥 ∈ 𝐵 𝜑 ) )