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Metamath Proof Explorer

Theorem supxrmnf2

Description: Removing minus infinity from a set does not affect its supremum. (Contributed by Glauco Siliprandi, 2-Jan-2022)

Ref Expression
Assertion supxrmnf2 ( 𝐴 ⊆ ℝ* → sup ( ( 𝐴 ∖ { -∞ } ) , ℝ* , < ) = sup ( 𝐴 , ℝ* , < ) )

Proof

Step Hyp Ref Expression
1 ssdifss ⊢ ( 𝐴 ⊆ ℝ* → ( 𝐴 ∖ { -∞ } ) ⊆ ℝ* )
2 supxrmnf ⊢ ( ( 𝐴 ∖ { -∞ } ) ⊆ ℝ* → sup ( ( ( 𝐴 ∖ { -∞ } ) ∪ { -∞ } ) , ℝ* , < ) = sup ( ( 𝐴 ∖ { -∞ } ) , ℝ* , < ) )
3 1 2 syl ⊢ ( 𝐴 ⊆ ℝ* → sup ( ( ( 𝐴 ∖ { -∞ } ) ∪ { -∞ } ) , ℝ* , < ) = sup ( ( 𝐴 ∖ { -∞ } ) , ℝ* , < ) )
4 3 adantr ⊢ ( ( 𝐴 ⊆ ℝ* ∧ -∞ ∈ 𝐴 ) → sup ( ( ( 𝐴 ∖ { -∞ } ) ∪ { -∞ } ) , ℝ* , < ) = sup ( ( 𝐴 ∖ { -∞ } ) , ℝ* , < ) )
5 difsnid ⊢ ( -∞ ∈ 𝐴 → ( ( 𝐴 ∖ { -∞ } ) ∪ { -∞ } ) = 𝐴 )
6 5 supeq1d ⊢ ( -∞ ∈ 𝐴 → sup ( ( ( 𝐴 ∖ { -∞ } ) ∪ { -∞ } ) , ℝ* , < ) = sup ( 𝐴 , ℝ* , < ) )
7 6 adantl ⊢ ( ( 𝐴 ⊆ ℝ* ∧ -∞ ∈ 𝐴 ) → sup ( ( ( 𝐴 ∖ { -∞ } ) ∪ { -∞ } ) , ℝ* , < ) = sup ( 𝐴 , ℝ* , < ) )
8 4 7 eqtr3d ⊢ ( ( 𝐴 ⊆ ℝ* ∧ -∞ ∈ 𝐴 ) → sup ( ( 𝐴 ∖ { -∞ } ) , ℝ* , < ) = sup ( 𝐴 , ℝ* , < ) )
9 difsn ⊢ ( ¬ -∞ ∈ 𝐴 → ( 𝐴 ∖ { -∞ } ) = 𝐴 )
10 9 supeq1d ⊢ ( ¬ -∞ ∈ 𝐴 → sup ( ( 𝐴 ∖ { -∞ } ) , ℝ* , < ) = sup ( 𝐴 , ℝ* , < ) )
11 10 adantl ⊢ ( ( 𝐴 ⊆ ℝ* ∧ ¬ -∞ ∈ 𝐴 ) → sup ( ( 𝐴 ∖ { -∞ } ) , ℝ* , < ) = sup ( 𝐴 , ℝ* , < ) )
12 8 11 pm2.61dan ⊢ ( 𝐴 ⊆ ℝ* → sup ( ( 𝐴 ∖ { -∞ } ) , ℝ* , < ) = sup ( 𝐴 , ℝ* , < ) )