Metamath Proof Explorer


Theorem orddif

Description: Ordinal derived from its successor. (Contributed by NM, 20-May-1998)

Ref Expression
Assertion orddif ( Ord 𝐴 → 𝐴 = ( suc 𝐴 ∖ { 𝐴 } ) )

Proof

Step Hyp Ref Expression
1 orddisj ⊢ ( Ord 𝐴 → ( 𝐴 ∩ { 𝐴 } ) = ∅ )
2 disj3 ⊢ ( ( 𝐴 ∩ { 𝐴 } ) = ∅ ↔ 𝐴 = ( 𝐴 ∖ { 𝐴 } ) )
3 df-suc ⊢ suc 𝐴 = ( 𝐴 ∪ { 𝐴 } )
4 3 difeq1i ⊢ ( suc 𝐴 ∖ { 𝐴 } ) = ( ( 𝐴 ∪ { 𝐴 } ) ∖ { 𝐴 } )
5 difun2 ⊢ ( ( 𝐴 ∪ { 𝐴 } ) ∖ { 𝐴 } ) = ( 𝐴 ∖ { 𝐴 } )
6 4 5 eqtri ⊢ ( suc 𝐴 ∖ { 𝐴 } ) = ( 𝐴 ∖ { 𝐴 } )
7 6 eqeq2i ⊢ ( 𝐴 = ( suc 𝐴 ∖ { 𝐴 } ) ↔ 𝐴 = ( 𝐴 ∖ { 𝐴 } ) )
8 2 7 bitr4i ⊢ ( ( 𝐴 ∩ { 𝐴 } ) = ∅ ↔ 𝐴 = ( suc 𝐴 ∖ { 𝐴 } ) )
9 1 8 sylib ⊢ ( Ord 𝐴 → 𝐴 = ( suc 𝐴 ∖ { 𝐴 } ) )