Metamath Proof Explorer


Theorem ttukeylem4

Description: Lemma for ttukey . (Contributed by Mario Carneiro, 15-May-2015)

Ref Expression
Hypotheses ttukeylem.1 ( 𝜑𝐹 : ( card ‘ ( 𝐴𝐵 ) ) –1-1-onto→ ( 𝐴𝐵 ) )
ttukeylem.2 ( 𝜑𝐵𝐴 )
ttukeylem.3 ( 𝜑 → ∀ 𝑥 ( 𝑥𝐴 ↔ ( 𝒫 𝑥 ∩ Fin ) ⊆ 𝐴 ) )
ttukeylem.4 𝐺 = recs ( ( 𝑧 ∈ V ↦ if ( dom 𝑧 = dom 𝑧 , if ( dom 𝑧 = ∅ , 𝐵 , ran 𝑧 ) , ( ( 𝑧 dom 𝑧 ) ∪ if ( ( ( 𝑧 dom 𝑧 ) ∪ { ( 𝐹 dom 𝑧 ) } ) ∈ 𝐴 , { ( 𝐹 dom 𝑧 ) } , ∅ ) ) ) ) )
Assertion ttukeylem4 ( 𝜑 → ( 𝐺 ‘ ∅ ) = 𝐵 )

Proof

Step Hyp Ref Expression
1 ttukeylem.1 ( 𝜑𝐹 : ( card ‘ ( 𝐴𝐵 ) ) –1-1-onto→ ( 𝐴𝐵 ) )
2 ttukeylem.2 ( 𝜑𝐵𝐴 )
3 ttukeylem.3 ( 𝜑 → ∀ 𝑥 ( 𝑥𝐴 ↔ ( 𝒫 𝑥 ∩ Fin ) ⊆ 𝐴 ) )
4 ttukeylem.4 𝐺 = recs ( ( 𝑧 ∈ V ↦ if ( dom 𝑧 = dom 𝑧 , if ( dom 𝑧 = ∅ , 𝐵 , ran 𝑧 ) , ( ( 𝑧 dom 𝑧 ) ∪ if ( ( ( 𝑧 dom 𝑧 ) ∪ { ( 𝐹 dom 𝑧 ) } ) ∈ 𝐴 , { ( 𝐹 dom 𝑧 ) } , ∅ ) ) ) ) )
5 0elon ∅ ∈ On
6 1 2 3 4 ttukeylem3 ( ( 𝜑 ∧ ∅ ∈ On ) → ( 𝐺 ‘ ∅ ) = if ( ∅ = ∅ , if ( ∅ = ∅ , 𝐵 , ( 𝐺 “ ∅ ) ) , ( ( 𝐺 ∅ ) ∪ if ( ( ( 𝐺 ∅ ) ∪ { ( 𝐹 ∅ ) } ) ∈ 𝐴 , { ( 𝐹 ∅ ) } , ∅ ) ) ) )
7 5 6 mpan2 ( 𝜑 → ( 𝐺 ‘ ∅ ) = if ( ∅ = ∅ , if ( ∅ = ∅ , 𝐵 , ( 𝐺 “ ∅ ) ) , ( ( 𝐺 ∅ ) ∪ if ( ( ( 𝐺 ∅ ) ∪ { ( 𝐹 ∅ ) } ) ∈ 𝐴 , { ( 𝐹 ∅ ) } , ∅ ) ) ) )
8 uni0 ∅ = ∅
9 8 eqcomi ∅ =
10 9 iftruei if ( ∅ = ∅ , if ( ∅ = ∅ , 𝐵 , ( 𝐺 “ ∅ ) ) , ( ( 𝐺 ∅ ) ∪ if ( ( ( 𝐺 ∅ ) ∪ { ( 𝐹 ∅ ) } ) ∈ 𝐴 , { ( 𝐹 ∅ ) } , ∅ ) ) ) = if ( ∅ = ∅ , 𝐵 , ( 𝐺 “ ∅ ) )
11 eqid ∅ = ∅
12 11 iftruei if ( ∅ = ∅ , 𝐵 , ( 𝐺 “ ∅ ) ) = 𝐵
13 10 12 eqtri if ( ∅ = ∅ , if ( ∅ = ∅ , 𝐵 , ( 𝐺 “ ∅ ) ) , ( ( 𝐺 ∅ ) ∪ if ( ( ( 𝐺 ∅ ) ∪ { ( 𝐹 ∅ ) } ) ∈ 𝐴 , { ( 𝐹 ∅ ) } , ∅ ) ) ) = 𝐵
14 7 13 syl6eq ( 𝜑 → ( 𝐺 ‘ ∅ ) = 𝐵 )