Metamath Proof Explorer


Theorem unabw

Description: Union of two class abstractions. Version of unab using implicit substitution, which does not require ax-8 , ax-10 , ax-12 . (Contributed by Gino Giotto, 15-Oct-2024)

Ref Expression
Hypotheses unabw.1 ( 𝑥 = 𝑦 → ( 𝜑𝜒 ) )
unabw.2 ( 𝑥 = 𝑦 → ( 𝜓𝜃 ) )
Assertion unabw ( { 𝑥𝜑 } ∪ { 𝑥𝜓 } ) = { 𝑦 ∣ ( 𝜒𝜃 ) }

Proof

Step Hyp Ref Expression
1 unabw.1 ( 𝑥 = 𝑦 → ( 𝜑𝜒 ) )
2 unabw.2 ( 𝑥 = 𝑦 → ( 𝜓𝜃 ) )
3 df-un ( { 𝑥𝜑 } ∪ { 𝑥𝜓 } ) = { 𝑦 ∣ ( 𝑦 ∈ { 𝑥𝜑 } ∨ 𝑦 ∈ { 𝑥𝜓 } ) }
4 df-clab ( 𝑦 ∈ { 𝑥𝜑 } ↔ [ 𝑦 / 𝑥 ] 𝜑 )
5 1 sbievw ( [ 𝑦 / 𝑥 ] 𝜑𝜒 )
6 4 5 bitri ( 𝑦 ∈ { 𝑥𝜑 } ↔ 𝜒 )
7 df-clab ( 𝑦 ∈ { 𝑥𝜓 } ↔ [ 𝑦 / 𝑥 ] 𝜓 )
8 2 sbievw ( [ 𝑦 / 𝑥 ] 𝜓𝜃 )
9 7 8 bitri ( 𝑦 ∈ { 𝑥𝜓 } ↔ 𝜃 )
10 6 9 orbi12i ( ( 𝑦 ∈ { 𝑥𝜑 } ∨ 𝑦 ∈ { 𝑥𝜓 } ) ↔ ( 𝜒𝜃 ) )
11 10 abbii { 𝑦 ∣ ( 𝑦 ∈ { 𝑥𝜑 } ∨ 𝑦 ∈ { 𝑥𝜓 } ) } = { 𝑦 ∣ ( 𝜒𝜃 ) }
12 3 11 eqtri ( { 𝑥𝜑 } ∪ { 𝑥𝜓 } ) = { 𝑦 ∣ ( 𝜒𝜃 ) }