Metamath Proof Explorer


Theorem dftpos5

Description: Alternate definition of tpos . (Contributed by Zhi Wang, 6-Oct-2025)

Ref Expression
Assertion dftpos5 tpos 𝐹 = ( 𝐹 ∘ ( ( 𝑥 dom 𝐹 { 𝑥 } ) ∪ { ⟨ ∅ , ∅ ⟩ } ) )

Proof

Step Hyp Ref Expression
1 df-tpos tpos 𝐹 = ( 𝐹 ∘ ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) )
2 mptun ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) = ( ( 𝑥 dom 𝐹 { 𝑥 } ) ∪ ( 𝑥 ∈ { ∅ } ↦ { 𝑥 } ) )
3 0ex ∅ ∈ V
4 sneq ( 𝑥 = ∅ → { 𝑥 } = { ∅ } )
5 4 cnveqd ( 𝑥 = ∅ → { 𝑥 } = { ∅ } )
6 5 unieqd ( 𝑥 = ∅ → { 𝑥 } = { ∅ } )
7 cnvsn0 { ∅ } = ∅
8 7 unieqi { ∅ } =
9 uni0 ∅ = ∅
10 8 9 eqtri { ∅ } = ∅
11 6 10 eqtrdi ( 𝑥 = ∅ → { 𝑥 } = ∅ )
12 11 fmptsng ( ( ∅ ∈ V ∧ ∅ ∈ V ) → { ⟨ ∅ , ∅ ⟩ } = ( 𝑥 ∈ { ∅ } ↦ { 𝑥 } ) )
13 3 3 12 mp2an { ⟨ ∅ , ∅ ⟩ } = ( 𝑥 ∈ { ∅ } ↦ { 𝑥 } )
14 13 uneq2i ( ( 𝑥 dom 𝐹 { 𝑥 } ) ∪ { ⟨ ∅ , ∅ ⟩ } ) = ( ( 𝑥 dom 𝐹 { 𝑥 } ) ∪ ( 𝑥 ∈ { ∅ } ↦ { 𝑥 } ) )
15 2 14 eqtr4i ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) = ( ( 𝑥 dom 𝐹 { 𝑥 } ) ∪ { ⟨ ∅ , ∅ ⟩ } )
16 15 coeq2i ( 𝐹 ∘ ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) ) = ( 𝐹 ∘ ( ( 𝑥 dom 𝐹 { 𝑥 } ) ∪ { ⟨ ∅ , ∅ ⟩ } ) )
17 1 16 eqtri tpos 𝐹 = ( 𝐹 ∘ ( ( 𝑥 dom 𝐹 { 𝑥 } ) ∪ { ⟨ ∅ , ∅ ⟩ } ) )