Metamath Proof Explorer


Theorem iinfssclem3

Description: Lemma for iinfssc . (Contributed by Zhi Wang, 31-Oct-2025)

Ref Expression
Hypotheses iinfssc.1 ⊢ ( 𝜑 → 𝐴 ≠ ∅ )
iinfssc.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐻 ⊆cat 𝐽 )
iinfssc.3 ⊢ ( 𝜑 → 𝐾 = ( 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 dom 𝐻 ↦ ∩ 𝑥 ∈ 𝐴 ( 𝐻 ‘ 𝑦 ) ) )
iinfssclem1.4 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝑆 = dom dom 𝐻 )
iinfssclem1.5 ⊢ Ⅎ 𝑥 𝜑
iinfssclem3.x ⊢ ( 𝜑 → 𝑋 ∈ ∩ 𝑥 ∈ 𝐴 𝑆 )
iinfssclem3.y ⊢ ( 𝜑 → 𝑌 ∈ ∩ 𝑥 ∈ 𝐴 𝑆 )
Assertion iinfssclem3 ( 𝜑 → ( 𝑋 𝐾 𝑌 ) = ∩ 𝑥 ∈ 𝐴 ( 𝑋 𝐻 𝑌 ) )

Proof

Step Hyp Ref Expression
1 iinfssc.1 ⊢ ( 𝜑 → 𝐴 ≠ ∅ )
2 iinfssc.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐻 ⊆cat 𝐽 )
3 iinfssc.3 ⊢ ( 𝜑 → 𝐾 = ( 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 dom 𝐻 ↦ ∩ 𝑥 ∈ 𝐴 ( 𝐻 ‘ 𝑦 ) ) )
4 iinfssclem1.4 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝑆 = dom dom 𝐻 )
5 iinfssclem1.5 ⊢ Ⅎ 𝑥 𝜑
6 iinfssclem3.x ⊢ ( 𝜑 → 𝑋 ∈ ∩ 𝑥 ∈ 𝐴 𝑆 )
7 iinfssclem3.y ⊢ ( 𝜑 → 𝑌 ∈ ∩ 𝑥 ∈ 𝐴 𝑆 )
8 1 2 3 4 5 iinfssclem1 ⊢ ( 𝜑 → 𝐾 = ( 𝑧 ∈ ∩ 𝑥 ∈ 𝐴 𝑆 , 𝑤 ∈ ∩ 𝑥 ∈ 𝐴 𝑆 ↦ ∩ 𝑥 ∈ 𝐴 ( 𝑧 𝐻 𝑤 ) ) )
9 nfv ⊢ Ⅎ 𝑥 ( 𝑧 = 𝑋 ∧ 𝑤 = 𝑌 )
10 5 9 nfan ⊢ Ⅎ 𝑥 ( 𝜑 ∧ ( 𝑧 = 𝑋 ∧ 𝑤 = 𝑌 ) )
11 simplrl ⊢ ( ( ( 𝜑 ∧ ( 𝑧 = 𝑋 ∧ 𝑤 = 𝑌 ) ) ∧ 𝑥 ∈ 𝐴 ) → 𝑧 = 𝑋 )
12 simplrr ⊢ ( ( ( 𝜑 ∧ ( 𝑧 = 𝑋 ∧ 𝑤 = 𝑌 ) ) ∧ 𝑥 ∈ 𝐴 ) → 𝑤 = 𝑌 )
13 11 12 oveq12d ⊢ ( ( ( 𝜑 ∧ ( 𝑧 = 𝑋 ∧ 𝑤 = 𝑌 ) ) ∧ 𝑥 ∈ 𝐴 ) → ( 𝑧 𝐻 𝑤 ) = ( 𝑋 𝐻 𝑌 ) )
14 10 13 iineq2d ⊢ ( ( 𝜑 ∧ ( 𝑧 = 𝑋 ∧ 𝑤 = 𝑌 ) ) → ∩ 𝑥 ∈ 𝐴 ( 𝑧 𝐻 𝑤 ) = ∩ 𝑥 ∈ 𝐴 ( 𝑋 𝐻 𝑌 ) )
15 ovex ⊢ ( 𝑋 𝐻 𝑌 ) ∈ V
16 15 rgenw ⊢ ∀ 𝑥 ∈ 𝐴 ( 𝑋 𝐻 𝑌 ) ∈ V
17 iinexg ⊢ ( ( 𝐴 ≠ ∅ ∧ ∀ 𝑥 ∈ 𝐴 ( 𝑋 𝐻 𝑌 ) ∈ V ) → ∩ 𝑥 ∈ 𝐴 ( 𝑋 𝐻 𝑌 ) ∈ V )
18 1 16 17 sylancl ⊢ ( 𝜑 → ∩ 𝑥 ∈ 𝐴 ( 𝑋 𝐻 𝑌 ) ∈ V )
19 8 14 6 7 18 ovmpod ⊢ ( 𝜑 → ( 𝑋 𝐾 𝑌 ) = ∩ 𝑥 ∈ 𝐴 ( 𝑋 𝐻 𝑌 ) )