Metamath Proof Explorer
Description: Equality deduction for Cartesian product. (Contributed by NM, 8-Dec-2013)
|
|
Ref |
Expression |
|
Hypotheses |
xpeq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
|
xpeq12d.2 |
⊢ ( 𝜑 → 𝐶 = 𝐷 ) |
|
Assertion |
xpeq12d |
⊢ ( 𝜑 → ( 𝐴 × 𝐶 ) = ( 𝐵 × 𝐷 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
xpeq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
| 2 |
|
xpeq12d.2 |
⊢ ( 𝜑 → 𝐶 = 𝐷 ) |
| 3 |
|
xpeq12 |
⊢ ( ( 𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ) → ( 𝐴 × 𝐶 ) = ( 𝐵 × 𝐷 ) ) |
| 4 |
1 2 3
|
syl2anc |
⊢ ( 𝜑 → ( 𝐴 × 𝐶 ) = ( 𝐵 × 𝐷 ) ) |