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Metamath Proof Explorer

Theorem oiiniseg

Description: ran F is an initial segment of A under the well-order R . (Contributed by Mario Carneiro, 26-Jun-2015)

Ref Expression
Hypothesis oicl.1 ⊢ 𝐹 = OrdIso ( 𝑅 , 𝐴 )
Assertion oiiniseg ( ( ( 𝑅 We 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝑁 ∈ 𝐴 ∧ 𝑀 ∈ dom 𝐹 ) ) → ( ( 𝐹 ‘ 𝑀 ) 𝑅 𝑁 ∨ 𝑁 ∈ ran 𝐹 ) )

Proof

Step Hyp Ref Expression
1 oicl.1 ⊢ 𝐹 = OrdIso ( 𝑅 , 𝐴 )
2 eqid ⊢ recs ( ( ℎ ∈ V ↦ ( ℩ 𝑣 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ∀ 𝑢 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ¬ 𝑢 𝑅 𝑣 ) ) ) = recs ( ( ℎ ∈ V ↦ ( ℩ 𝑣 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ∀ 𝑢 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ¬ 𝑢 𝑅 𝑣 ) ) )
3 eqid ⊢ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } = { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 }
4 eqid ⊢ ( ℎ ∈ V ↦ ( ℩ 𝑣 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ∀ 𝑢 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ¬ 𝑢 𝑅 𝑣 ) ) = ( ℎ ∈ V ↦ ( ℩ 𝑣 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ∀ 𝑢 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ¬ 𝑢 𝑅 𝑣 ) )
5 2 3 4 ordtypecbv ⊢ recs ( ( 𝑓 ∈ V ↦ ( ℩ 𝑠 ∈ { 𝑦 ∈ 𝐴 ∣ ∀ 𝑖 ∈ ran 𝑓 𝑖 𝑅 𝑦 } ∀ 𝑟 ∈ { 𝑦 ∈ 𝐴 ∣ ∀ 𝑖 ∈ ran 𝑓 𝑖 𝑅 𝑦 } ¬ 𝑟 𝑅 𝑠 ) ) ) = recs ( ( ℎ ∈ V ↦ ( ℩ 𝑣 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ∀ 𝑢 ∈ { 𝑤 ∈ 𝐴 ∣ ∀ 𝑗 ∈ ran ℎ 𝑗 𝑅 𝑤 } ¬ 𝑢 𝑅 𝑣 ) ) )
6 eqid ⊢ { 𝑥 ∈ On ∣ ∃ 𝑡 ∈ 𝐴 ∀ 𝑧 ∈ ( recs ( ( 𝑓 ∈ V ↦ ( ℩ 𝑠 ∈ { 𝑦 ∈ 𝐴 ∣ ∀ 𝑖 ∈ ran 𝑓 𝑖 𝑅 𝑦 } ∀ 𝑟 ∈ { 𝑦 ∈ 𝐴 ∣ ∀ 𝑖 ∈ ran 𝑓 𝑖 𝑅 𝑦 } ¬ 𝑟 𝑅 𝑠 ) ) ) “ 𝑥 ) 𝑧 𝑅 𝑡 } = { 𝑥 ∈ On ∣ ∃ 𝑡 ∈ 𝐴 ∀ 𝑧 ∈ ( recs ( ( 𝑓 ∈ V ↦ ( ℩ 𝑠 ∈ { 𝑦 ∈ 𝐴 ∣ ∀ 𝑖 ∈ ran 𝑓 𝑖 𝑅 𝑦 } ∀ 𝑟 ∈ { 𝑦 ∈ 𝐴 ∣ ∀ 𝑖 ∈ ran 𝑓 𝑖 𝑅 𝑦 } ¬ 𝑟 𝑅 𝑠 ) ) ) “ 𝑥 ) 𝑧 𝑅 𝑡 }
7 simpl ⊢ ( ( 𝑅 We 𝐴 ∧ 𝑅 Se 𝐴 ) → 𝑅 We 𝐴 )
8 simpr ⊢ ( ( 𝑅 We 𝐴 ∧ 𝑅 Se 𝐴 ) → 𝑅 Se 𝐴 )
9 5 3 4 6 1 7 8 ordtypelem7 ⊢ ( ( ( ( 𝑅 We 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ 𝑁 ∈ 𝐴 ) ∧ 𝑀 ∈ dom 𝐹 ) → ( ( 𝐹 ‘ 𝑀 ) 𝑅 𝑁 ∨ 𝑁 ∈ ran 𝐹 ) )
10 9 anasss ⊢ ( ( ( 𝑅 We 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝑁 ∈ 𝐴 ∧ 𝑀 ∈ dom 𝐹 ) ) → ( ( 𝐹 ‘ 𝑀 ) 𝑅 𝑁 ∨ 𝑁 ∈ ran 𝐹 ) )