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Description: The sets of a cartesian product are disjoint if the sets in the first argument are disjoint. (Contributed by Glauco Siliprandi, 11-Oct-2020)
| Ref | Expression | ||
|---|---|---|---|
| Hypothesis | disjxp1.1 | ⊢ ( 𝜑 → Disj 𝑥 ∈ 𝐴 𝐵 ) | |
| Assertion | disjxp1 | ⊢ ( 𝜑 → Disj 𝑥 ∈ 𝐴 ( 𝐵 × 𝐶 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | disjxp1.1 | ⊢ ( 𝜑 → Disj 𝑥 ∈ 𝐴 𝐵 ) | |
| 2 | animorrl | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 = 𝑧 ) → ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ∩ ⦋ 𝑧 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ) = ∅ ) ) | |
| 3 | csbxp | ⊢ ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) = ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 × ⦋ 𝑦 / 𝑥 ⦌ 𝐶 ) | |
| 4 | csbxp | ⊢ ⦋ 𝑧 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) = ( ⦋ 𝑧 / 𝑥 ⦌ 𝐵 × ⦋ 𝑧 / 𝑥 ⦌ 𝐶 ) | |
| 5 | 3 4 | ineq12i | ⊢ ( ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ∩ ⦋ 𝑧 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ) = ( ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 × ⦋ 𝑦 / 𝑥 ⦌ 𝐶 ) ∩ ( ⦋ 𝑧 / 𝑥 ⦌ 𝐵 × ⦋ 𝑧 / 𝑥 ⦌ 𝐶 ) ) |
| 6 | simpll | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → 𝜑 ) | |
| 7 | simplrl | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → 𝑦 ∈ 𝐴 ) | |
| 8 | simplrr | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → 𝑧 ∈ 𝐴 ) | |
| 9 | 6 7 8 | jca31 | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → ( ( 𝜑 ∧ 𝑦 ∈ 𝐴 ) ∧ 𝑧 ∈ 𝐴 ) ) |
| 10 | simpr | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → 𝑦 ≠ 𝑧 ) | |
| 11 | 10 | neneqd | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → ¬ 𝑦 = 𝑧 ) |
| 12 | disjors | ⊢ ( Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∩ ⦋ 𝑧 / 𝑥 ⦌ 𝐵 ) = ∅ ) ) | |
| 13 | 1 12 | sylib | ⊢ ( 𝜑 → ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∩ ⦋ 𝑧 / 𝑥 ⦌ 𝐵 ) = ∅ ) ) |
| 14 | 13 | r19.21bi | ⊢ ( ( 𝜑 ∧ 𝑦 ∈ 𝐴 ) → ∀ 𝑧 ∈ 𝐴 ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∩ ⦋ 𝑧 / 𝑥 ⦌ 𝐵 ) = ∅ ) ) |
| 15 | 14 | r19.21bi | ⊢ ( ( ( 𝜑 ∧ 𝑦 ∈ 𝐴 ) ∧ 𝑧 ∈ 𝐴 ) → ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∩ ⦋ 𝑧 / 𝑥 ⦌ 𝐵 ) = ∅ ) ) |
| 16 | 15 | ord | ⊢ ( ( ( 𝜑 ∧ 𝑦 ∈ 𝐴 ) ∧ 𝑧 ∈ 𝐴 ) → ( ¬ 𝑦 = 𝑧 → ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∩ ⦋ 𝑧 / 𝑥 ⦌ 𝐵 ) = ∅ ) ) |
| 17 | 9 11 16 | sylc | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∩ ⦋ 𝑧 / 𝑥 ⦌ 𝐵 ) = ∅ ) |
| 18 | xpdisj1 | ⊢ ( ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∩ ⦋ 𝑧 / 𝑥 ⦌ 𝐵 ) = ∅ → ( ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 × ⦋ 𝑦 / 𝑥 ⦌ 𝐶 ) ∩ ( ⦋ 𝑧 / 𝑥 ⦌ 𝐵 × ⦋ 𝑧 / 𝑥 ⦌ 𝐶 ) ) = ∅ ) | |
| 19 | 17 18 | syl | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → ( ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 × ⦋ 𝑦 / 𝑥 ⦌ 𝐶 ) ∩ ( ⦋ 𝑧 / 𝑥 ⦌ 𝐵 × ⦋ 𝑧 / 𝑥 ⦌ 𝐶 ) ) = ∅ ) |
| 20 | 5 19 | eqtrid | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → ( ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ∩ ⦋ 𝑧 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ) = ∅ ) |
| 21 | 20 | olcd | ⊢ ( ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) ∧ 𝑦 ≠ 𝑧 ) → ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ∩ ⦋ 𝑧 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ) = ∅ ) ) |
| 22 | 2 21 | pm2.61dane | ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) → ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ∩ ⦋ 𝑧 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ) = ∅ ) ) |
| 23 | 22 | ralrimivva | ⊢ ( 𝜑 → ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ∩ ⦋ 𝑧 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ) = ∅ ) ) |
| 24 | disjors | ⊢ ( Disj 𝑥 ∈ 𝐴 ( 𝐵 × 𝐶 ) ↔ ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( 𝑦 = 𝑧 ∨ ( ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ∩ ⦋ 𝑧 / 𝑥 ⦌ ( 𝐵 × 𝐶 ) ) = ∅ ) ) | |
| 25 | 23 24 | sylibr | ⊢ ( 𝜑 → Disj 𝑥 ∈ 𝐴 ( 𝐵 × 𝐶 ) ) |