Metamath Proof Explorer


Theorem finxpnom

Description: Cartesian exponentiation when the exponent is not a natural number defaults to the empty set. (Contributed by ML, 24-Oct-2020)

Ref Expression
Assertion finxpnom ( ¬ 𝑁 ∈ ω → ( 𝑈 ↑↑ 𝑁 ) = ∅ )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝑁 ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ 𝑁 , 𝑦 ⟩ ) ‘ 𝑁 ) ) → 𝑁 ∈ ω )
2 1 con3i ( ¬ 𝑁 ∈ ω → ¬ ( 𝑁 ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ 𝑁 , 𝑦 ⟩ ) ‘ 𝑁 ) ) )
3 abid ( 𝑦 ∈ { 𝑦 ∣ ( 𝑁 ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ 𝑁 , 𝑦 ⟩ ) ‘ 𝑁 ) ) } ↔ ( 𝑁 ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ 𝑁 , 𝑦 ⟩ ) ‘ 𝑁 ) ) )
4 2 3 sylnibr ( ¬ 𝑁 ∈ ω → ¬ 𝑦 ∈ { 𝑦 ∣ ( 𝑁 ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ 𝑁 , 𝑦 ⟩ ) ‘ 𝑁 ) ) } )
5 df-finxp ( 𝑈 ↑↑ 𝑁 ) = { 𝑦 ∣ ( 𝑁 ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ 𝑁 , 𝑦 ⟩ ) ‘ 𝑁 ) ) }
6 5 eleq2i ( 𝑦 ∈ ( 𝑈 ↑↑ 𝑁 ) ↔ 𝑦 ∈ { 𝑦 ∣ ( 𝑁 ∈ ω ∧ ∅ = ( rec ( ( 𝑛 ∈ ω , 𝑥 ∈ V ↦ if ( ( 𝑛 = 1o𝑥𝑈 ) , ∅ , if ( 𝑥 ∈ ( V × 𝑈 ) , ⟨ 𝑛 , ( 1st𝑥 ) ⟩ , ⟨ 𝑛 , 𝑥 ⟩ ) ) ) , ⟨ 𝑁 , 𝑦 ⟩ ) ‘ 𝑁 ) ) } )
7 4 6 sylnibr ( ¬ 𝑁 ∈ ω → ¬ 𝑦 ∈ ( 𝑈 ↑↑ 𝑁 ) )
8 7 eq0rdv ( ¬ 𝑁 ∈ ω → ( 𝑈 ↑↑ 𝑁 ) = ∅ )