Metamath Proof Explorer


Theorem bj-rest10

Description: An elementwise intersection on a nonempty family by the empty set is the singleton on the empty set. TODO: this generalizes rest0 and could replace it. (Contributed by BJ, 27-Apr-2021)

Ref Expression
Assertion bj-rest10 ( 𝑋𝑉 → ( 𝑋 ≠ ∅ → ( 𝑋t ∅ ) = { ∅ } ) )

Proof

Step Hyp Ref Expression
1 0ex ∅ ∈ V
2 elrest ( ( 𝑋𝑉 ∧ ∅ ∈ V ) → ( 𝑥 ∈ ( 𝑋t ∅ ) ↔ ∃ 𝑦𝑋 𝑥 = ( 𝑦 ∩ ∅ ) ) )
3 1 2 mpan2 ( 𝑋𝑉 → ( 𝑥 ∈ ( 𝑋t ∅ ) ↔ ∃ 𝑦𝑋 𝑥 = ( 𝑦 ∩ ∅ ) ) )
4 in0 ( 𝑦 ∩ ∅ ) = ∅
5 4 eqeq2i ( 𝑥 = ( 𝑦 ∩ ∅ ) ↔ 𝑥 = ∅ )
6 5 rexbii ( ∃ 𝑦𝑋 𝑥 = ( 𝑦 ∩ ∅ ) ↔ ∃ 𝑦𝑋 𝑥 = ∅ )
7 df-rex ( ∃ 𝑦𝑋 𝑥 = ∅ ↔ ∃ 𝑦 ( 𝑦𝑋𝑥 = ∅ ) )
8 19.41v ( ∃ 𝑦 ( 𝑦𝑋𝑥 = ∅ ) ↔ ( ∃ 𝑦 𝑦𝑋𝑥 = ∅ ) )
9 n0 ( 𝑋 ≠ ∅ ↔ ∃ 𝑦 𝑦𝑋 )
10 9 bicomi ( ∃ 𝑦 𝑦𝑋𝑋 ≠ ∅ )
11 10 anbi1i ( ( ∃ 𝑦 𝑦𝑋𝑥 = ∅ ) ↔ ( 𝑋 ≠ ∅ ∧ 𝑥 = ∅ ) )
12 8 11 bitri ( ∃ 𝑦 ( 𝑦𝑋𝑥 = ∅ ) ↔ ( 𝑋 ≠ ∅ ∧ 𝑥 = ∅ ) )
13 7 12 bitri ( ∃ 𝑦𝑋 𝑥 = ∅ ↔ ( 𝑋 ≠ ∅ ∧ 𝑥 = ∅ ) )
14 6 13 bitri ( ∃ 𝑦𝑋 𝑥 = ( 𝑦 ∩ ∅ ) ↔ ( 𝑋 ≠ ∅ ∧ 𝑥 = ∅ ) )
15 14 baib ( 𝑋 ≠ ∅ → ( ∃ 𝑦𝑋 𝑥 = ( 𝑦 ∩ ∅ ) ↔ 𝑥 = ∅ ) )
16 3 15 sylan9bb ( ( 𝑋𝑉𝑋 ≠ ∅ ) → ( 𝑥 ∈ ( 𝑋t ∅ ) ↔ 𝑥 = ∅ ) )
17 velsn ( 𝑥 ∈ { ∅ } ↔ 𝑥 = ∅ )
18 16 17 bitr4di ( ( 𝑋𝑉𝑋 ≠ ∅ ) → ( 𝑥 ∈ ( 𝑋t ∅ ) ↔ 𝑥 ∈ { ∅ } ) )
19 18 eqrdv ( ( 𝑋𝑉𝑋 ≠ ∅ ) → ( 𝑋t ∅ ) = { ∅ } )
20 19 ex ( 𝑋𝑉 → ( 𝑋 ≠ ∅ → ( 𝑋t ∅ ) = { ∅ } ) )