Description: Subset theorem for Cartesian product. Generalization of Theorem 101 of Suppes p. 52. (Contributed by NM, 26-Aug-1995) (Proof shortened by Andrew Salmon, 27-Aug-2011)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | xpss12 | ⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷 ) → ( 𝐴 × 𝐶 ) ⊆ ( 𝐵 × 𝐷 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | ssel | ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ) | |
| 2 | ssel | ⊢ ( 𝐶 ⊆ 𝐷 → ( 𝑦 ∈ 𝐶 → 𝑦 ∈ 𝐷 ) ) | |
| 3 | 1 2 | im2anan9 | ⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷 ) → ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶 ) → ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐷 ) ) ) |
| 4 | 3 | ssopab2dv | ⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷 ) → { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶 ) } ⊆ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐷 ) } ) |
| 5 | df-xp | ⊢ ( 𝐴 × 𝐶 ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶 ) } | |
| 6 | df-xp | ⊢ ( 𝐵 × 𝐷 ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐷 ) } | |
| 7 | 4 5 6 | 3sstr4g | ⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷 ) → ( 𝐴 × 𝐶 ) ⊆ ( 𝐵 × 𝐷 ) ) |