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Metamath Proof Explorer

Theorem rabss2OLD

Description: Obsolete version of rabss2 as of 1-Feb-2026. (Contributed by NM, 30-May-2006) (Proof shortened by Andrew Salmon, 26-Jun-2011) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion rabss2OLD ( 𝐴 ⊆ 𝐵 → { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐵 ∣ 𝜑 } )

Proof

Step Hyp Ref Expression
1 pm3.45 ⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
2 1 alimi ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
3 df-ss ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
4 ss2ab ⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } ↔ ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
5 2 3 4 3imtr4i ⊢ ( 𝐴 ⊆ 𝐵 → { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ⊆ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } )
6 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
7 df-rab ⊢ { 𝑥 ∈ 𝐵 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) }
8 5 6 7 3sstr4g ⊢ ( 𝐴 ⊆ 𝐵 → { 𝑥 ∈ 𝐴 ∣ 𝜑 } ⊆ { 𝑥 ∈ 𝐵 ∣ 𝜑 } )