Metamath Proof Explorer


Theorem infsubc

Description: The intersection of two subcategories is a subcategory. (Contributed by Zhi Wang, 31-Oct-2025)

Ref Expression
Assertion infsubc ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) → ( 𝑥 ∈ ( dom 𝐴 ∩ dom 𝐵 ) ↦ ( ( 𝐴𝑥 ) ∩ ( 𝐵𝑥 ) ) ) ∈ ( Subcat ‘ 𝐶 ) )

Proof

Step Hyp Ref Expression
1 prnzg ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) → { 𝐴 , 𝐵 } ≠ ∅ )
2 1 adantr ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) → { 𝐴 , 𝐵 } ≠ ∅ )
3 simpll ( ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) ∧ 𝑦 ∈ { 𝐴 , 𝐵 } ) → 𝐴 ∈ ( Subcat ‘ 𝐶 ) )
4 eleq1 ( 𝑦 = 𝐴 → ( 𝑦 ∈ ( Subcat ‘ 𝐶 ) ↔ 𝐴 ∈ ( Subcat ‘ 𝐶 ) ) )
5 3 4 syl5ibrcom ( ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) ∧ 𝑦 ∈ { 𝐴 , 𝐵 } ) → ( 𝑦 = 𝐴𝑦 ∈ ( Subcat ‘ 𝐶 ) ) )
6 simplr ( ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) ∧ 𝑦 ∈ { 𝐴 , 𝐵 } ) → 𝐵 ∈ ( Subcat ‘ 𝐶 ) )
7 eleq1 ( 𝑦 = 𝐵 → ( 𝑦 ∈ ( Subcat ‘ 𝐶 ) ↔ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) )
8 6 7 syl5ibrcom ( ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) ∧ 𝑦 ∈ { 𝐴 , 𝐵 } ) → ( 𝑦 = 𝐵𝑦 ∈ ( Subcat ‘ 𝐶 ) ) )
9 elpri ( 𝑦 ∈ { 𝐴 , 𝐵 } → ( 𝑦 = 𝐴𝑦 = 𝐵 ) )
10 9 adantl ( ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) ∧ 𝑦 ∈ { 𝐴 , 𝐵 } ) → ( 𝑦 = 𝐴𝑦 = 𝐵 ) )
11 5 8 10 mpjaod ( ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) ∧ 𝑦 ∈ { 𝐴 , 𝐵 } ) → 𝑦 ∈ ( Subcat ‘ 𝐶 ) )
12 iinfprg ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) → ( 𝑥 ∈ ( dom 𝐴 ∩ dom 𝐵 ) ↦ ( ( 𝐴𝑥 ) ∩ ( 𝐵𝑥 ) ) ) = ( 𝑥 𝑦 ∈ { 𝐴 , 𝐵 } dom 𝑦 𝑦 ∈ { 𝐴 , 𝐵 } ( 𝑦𝑥 ) ) )
13 2 11 12 iinfsubc ( ( 𝐴 ∈ ( Subcat ‘ 𝐶 ) ∧ 𝐵 ∈ ( Subcat ‘ 𝐶 ) ) → ( 𝑥 ∈ ( dom 𝐴 ∩ dom 𝐵 ) ↦ ( ( 𝐴𝑥 ) ∩ ( 𝐵𝑥 ) ) ) ∈ ( Subcat ‘ 𝐶 ) )