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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 GG, 15-Oct-2024)

		Ref	Expression
	Hypotheses	unabw.1	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜒 ) )
		unabw.2	⊢ ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜃 ) )
	Assertion	unabw	⊢ ( { 𝑥 ∣ 𝜑 } ∪ { 𝑥 ∣ 𝜓 } ) = { 𝑦 ∣ ( 𝜒 ∨ 𝜃 ) }

Proof

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