Metamath Proof Explorer


Theorem ordtypelem1

Description: Lemma for ordtype . (Contributed by Mario Carneiro, 24-Jun-2015)

Ref Expression
Hypotheses ordtypelem.1 𝐹 = recs ( 𝐺 )
ordtypelem.2 𝐶 = { 𝑤𝐴 ∣ ∀ 𝑗 ∈ ran 𝑗 𝑅 𝑤 }
ordtypelem.3 𝐺 = ( ∈ V ↦ ( 𝑣𝐶𝑢𝐶 ¬ 𝑢 𝑅 𝑣 ) )
ordtypelem.5 𝑇 = { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 }
ordtypelem.6 𝑂 = OrdIso ( 𝑅 , 𝐴 )
ordtypelem.7 ( 𝜑𝑅 We 𝐴 )
ordtypelem.8 ( 𝜑𝑅 Se 𝐴 )
Assertion ordtypelem1 ( 𝜑𝑂 = ( 𝐹𝑇 ) )

Proof

Step Hyp Ref Expression
1 ordtypelem.1 𝐹 = recs ( 𝐺 )
2 ordtypelem.2 𝐶 = { 𝑤𝐴 ∣ ∀ 𝑗 ∈ ran 𝑗 𝑅 𝑤 }
3 ordtypelem.3 𝐺 = ( ∈ V ↦ ( 𝑣𝐶𝑢𝐶 ¬ 𝑢 𝑅 𝑣 ) )
4 ordtypelem.5 𝑇 = { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 }
5 ordtypelem.6 𝑂 = OrdIso ( 𝑅 , 𝐴 )
6 ordtypelem.7 ( 𝜑𝑅 We 𝐴 )
7 ordtypelem.8 ( 𝜑𝑅 Se 𝐴 )
8 iftrue ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) → if ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) , ( 𝐹 ↾ { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 } ) , ∅ ) = ( 𝐹 ↾ { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 } ) )
9 6 7 8 syl2anc ( 𝜑 → if ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) , ( 𝐹 ↾ { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 } ) , ∅ ) = ( 𝐹 ↾ { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 } ) )
10 2 3 1 dfoi OrdIso ( 𝑅 , 𝐴 ) = if ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) , ( 𝐹 ↾ { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 } ) , ∅ )
11 5 10 eqtri 𝑂 = if ( ( 𝑅 We 𝐴𝑅 Se 𝐴 ) , ( 𝐹 ↾ { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 } ) , ∅ )
12 4 reseq2i ( 𝐹𝑇 ) = ( 𝐹 ↾ { 𝑥 ∈ On ∣ ∃ 𝑡𝐴𝑧 ∈ ( 𝐹𝑥 ) 𝑧 𝑅 𝑡 } )
13 9 11 12 3eqtr4g ( 𝜑𝑂 = ( 𝐹𝑇 ) )