Metamath Proof Explorer


Theorem finxp0

Description: The value of Cartesian exponentiation at zero. (Contributed by ML, 24-Oct-2020)

Ref Expression
Assertion finxp0 ( 𝑈 ↑↑ ∅ ) = ∅

Proof

Step Hyp Ref Expression
1 0ex ∅ ∈ V
2 vex 𝑦 ∈ V
3 1 2 opnzi ⟨ ∅ , 𝑦 ⟩ ≠ ∅
4 3 nesymi ¬ ∅ = ⟨ ∅ , 𝑦
5 peano1 ∅ ∈ ω
6 df-finxp ( 𝑈 ↑↑ ∅ ) = { 𝑦 ∣ ( ∅ ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ ∅ , 𝑦 ⟩ ) ‘ ∅ ) ) }
7 6 abeq2i ( 𝑦 ∈ ( 𝑈 ↑↑ ∅ ) ↔ ( ∅ ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ ∅ , 𝑦 ⟩ ) ‘ ∅ ) ) )
8 5 7 mpbiran ( 𝑦 ∈ ( 𝑈 ↑↑ ∅ ) ↔ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ ∅ , 𝑦 ⟩ ) ‘ ∅ ) )
9 opex ⟨ ∅ , 𝑦 ⟩ ∈ V
10 9 rdg0 ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ ∅ , 𝑦 ⟩ ) ‘ ∅ ) = ⟨ ∅ , 𝑦
11 10 eqeq2i ( ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ ∅ , 𝑦 ⟩ ) ‘ ∅ ) ↔ ∅ = ⟨ ∅ , 𝑦 ⟩ )
12 8 11 bitri ( 𝑦 ∈ ( 𝑈 ↑↑ ∅ ) ↔ ∅ = ⟨ ∅ , 𝑦 ⟩ )
13 4 12 mtbir ¬ 𝑦 ∈ ( 𝑈 ↑↑ ∅ )
14 13 nel0 ( 𝑈 ↑↑ ∅ ) = ∅