Metamath Proof Explorer


Theorem bdaybndbday

Description: Bounds formed from the birthday have the same birthday. (Contributed by RP, 30-Sep-2023)

Ref Expression
Assertion bdaybndbday ( ( 𝐴 No 𝐵 = ( bday 𝐴 ) ∧ 𝐶 ∈ { 1o , 2o } ) → ( bday ‘ ( 𝐵 × { 𝐶 } ) ) = ( bday 𝐴 ) )

Proof

Step Hyp Ref Expression
1 bdaybndex ( ( 𝐴 No 𝐵 = ( bday 𝐴 ) ∧ 𝐶 ∈ { 1o , 2o } ) → ( 𝐵 × { 𝐶 } ) ∈ No )
2 bdayval ( ( 𝐵 × { 𝐶 } ) ∈ No → ( bday ‘ ( 𝐵 × { 𝐶 } ) ) = dom ( 𝐵 × { 𝐶 } ) )
3 1 2 syl ( ( 𝐴 No 𝐵 = ( bday 𝐴 ) ∧ 𝐶 ∈ { 1o , 2o } ) → ( bday ‘ ( 𝐵 × { 𝐶 } ) ) = dom ( 𝐵 × { 𝐶 } ) )
4 simp3 ( ( 𝐴 No 𝐵 = ( bday 𝐴 ) ∧ 𝐶 ∈ { 1o , 2o } ) → 𝐶 ∈ { 1o , 2o } )
5 snnzg ( 𝐶 ∈ { 1o , 2o } → { 𝐶 } ≠ ∅ )
6 dmxp ( { 𝐶 } ≠ ∅ → dom ( 𝐵 × { 𝐶 } ) = 𝐵 )
7 4 5 6 3syl ( ( 𝐴 No 𝐵 = ( bday 𝐴 ) ∧ 𝐶 ∈ { 1o , 2o } ) → dom ( 𝐵 × { 𝐶 } ) = 𝐵 )
8 simp2 ( ( 𝐴 No 𝐵 = ( bday 𝐴 ) ∧ 𝐶 ∈ { 1o , 2o } ) → 𝐵 = ( bday 𝐴 ) )
9 3 7 8 3eqtrd ( ( 𝐴 No 𝐵 = ( bday 𝐴 ) ∧ 𝐶 ∈ { 1o , 2o } ) → ( bday ‘ ( 𝐵 × { 𝐶 } ) ) = ( bday 𝐴 ) )