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Description: Equivalence relation for positive fractions in terms of positive integers. (Contributed by Mario Carneiro, 8-May-2013) (New usage is discouraged.)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | enqbreq2 | ⊢ ( ( 𝐴 ∈ ( N × N ) ∧ 𝐵 ∈ ( N × N ) ) → ( 𝐴 ~Q 𝐵 ↔ ( ( 1st ‘ 𝐴 ) ·N ( 2nd ‘ 𝐵 ) ) = ( ( 1st ‘ 𝐵 ) ·N ( 2nd ‘ 𝐴 ) ) ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 1st2nd2 | ⊢ ( 𝐴 ∈ ( N × N ) → 𝐴 = 〈 ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) 〉 ) | |
| 2 | 1st2nd2 | ⊢ ( 𝐵 ∈ ( N × N ) → 𝐵 = 〈 ( 1st ‘ 𝐵 ) , ( 2nd ‘ 𝐵 ) 〉 ) | |
| 3 | 1 2 | breqan12d | ⊢ ( ( 𝐴 ∈ ( N × N ) ∧ 𝐵 ∈ ( N × N ) ) → ( 𝐴 ~Q 𝐵 ↔ 〈 ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) 〉 ~Q 〈 ( 1st ‘ 𝐵 ) , ( 2nd ‘ 𝐵 ) 〉 ) ) |
| 4 | xp1st | ⊢ ( 𝐴 ∈ ( N × N ) → ( 1st ‘ 𝐴 ) ∈ N ) | |
| 5 | xp2nd | ⊢ ( 𝐴 ∈ ( N × N ) → ( 2nd ‘ 𝐴 ) ∈ N ) | |
| 6 | 4 5 | jca | ⊢ ( 𝐴 ∈ ( N × N ) → ( ( 1st ‘ 𝐴 ) ∈ N ∧ ( 2nd ‘ 𝐴 ) ∈ N ) ) |
| 7 | xp1st | ⊢ ( 𝐵 ∈ ( N × N ) → ( 1st ‘ 𝐵 ) ∈ N ) | |
| 8 | xp2nd | ⊢ ( 𝐵 ∈ ( N × N ) → ( 2nd ‘ 𝐵 ) ∈ N ) | |
| 9 | 7 8 | jca | ⊢ ( 𝐵 ∈ ( N × N ) → ( ( 1st ‘ 𝐵 ) ∈ N ∧ ( 2nd ‘ 𝐵 ) ∈ N ) ) |
| 10 | enqbreq | ⊢ ( ( ( ( 1st ‘ 𝐴 ) ∈ N ∧ ( 2nd ‘ 𝐴 ) ∈ N ) ∧ ( ( 1st ‘ 𝐵 ) ∈ N ∧ ( 2nd ‘ 𝐵 ) ∈ N ) ) → ( 〈 ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) 〉 ~Q 〈 ( 1st ‘ 𝐵 ) , ( 2nd ‘ 𝐵 ) 〉 ↔ ( ( 1st ‘ 𝐴 ) ·N ( 2nd ‘ 𝐵 ) ) = ( ( 2nd ‘ 𝐴 ) ·N ( 1st ‘ 𝐵 ) ) ) ) | |
| 11 | 6 9 10 | syl2an | ⊢ ( ( 𝐴 ∈ ( N × N ) ∧ 𝐵 ∈ ( N × N ) ) → ( 〈 ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) 〉 ~Q 〈 ( 1st ‘ 𝐵 ) , ( 2nd ‘ 𝐵 ) 〉 ↔ ( ( 1st ‘ 𝐴 ) ·N ( 2nd ‘ 𝐵 ) ) = ( ( 2nd ‘ 𝐴 ) ·N ( 1st ‘ 𝐵 ) ) ) ) |
| 12 | mulcompi | ⊢ ( ( 2nd ‘ 𝐴 ) ·N ( 1st ‘ 𝐵 ) ) = ( ( 1st ‘ 𝐵 ) ·N ( 2nd ‘ 𝐴 ) ) | |
| 13 | 12 | eqeq2i | ⊢ ( ( ( 1st ‘ 𝐴 ) ·N ( 2nd ‘ 𝐵 ) ) = ( ( 2nd ‘ 𝐴 ) ·N ( 1st ‘ 𝐵 ) ) ↔ ( ( 1st ‘ 𝐴 ) ·N ( 2nd ‘ 𝐵 ) ) = ( ( 1st ‘ 𝐵 ) ·N ( 2nd ‘ 𝐴 ) ) ) |
| 14 | 13 | a1i | ⊢ ( ( 𝐴 ∈ ( N × N ) ∧ 𝐵 ∈ ( N × N ) ) → ( ( ( 1st ‘ 𝐴 ) ·N ( 2nd ‘ 𝐵 ) ) = ( ( 2nd ‘ 𝐴 ) ·N ( 1st ‘ 𝐵 ) ) ↔ ( ( 1st ‘ 𝐴 ) ·N ( 2nd ‘ 𝐵 ) ) = ( ( 1st ‘ 𝐵 ) ·N ( 2nd ‘ 𝐴 ) ) ) ) |
| 15 | 3 11 14 | 3bitrd | ⊢ ( ( 𝐴 ∈ ( N × N ) ∧ 𝐵 ∈ ( N × N ) ) → ( 𝐴 ~Q 𝐵 ↔ ( ( 1st ‘ 𝐴 ) ·N ( 2nd ‘ 𝐵 ) ) = ( ( 1st ‘ 𝐵 ) ·N ( 2nd ‘ 𝐴 ) ) ) ) |