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Metamath Proof Explorer

Theorem ind1a

Description: Value of the indicator function where it is 1 . (Contributed by Thierry Arnoux, 22-Aug-2017)

Ref Expression
Assertion ind1a ( ( 𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂 ∧ 𝑋 ∈ 𝑂 ) → ( ( ( ( 𝟭 ‘ 𝑂 ) ‘ 𝐴 ) ‘ 𝑋 ) = 1 ↔ 𝑋 ∈ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 indfval ⊢ ( ( 𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂 ∧ 𝑋 ∈ 𝑂 ) → ( ( ( 𝟭 ‘ 𝑂 ) ‘ 𝐴 ) ‘ 𝑋 ) = if ( 𝑋 ∈ 𝐴 , 1 , 0 ) )
2 1 eqeq1d ⊢ ( ( 𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂 ∧ 𝑋 ∈ 𝑂 ) → ( ( ( ( 𝟭 ‘ 𝑂 ) ‘ 𝐴 ) ‘ 𝑋 ) = 1 ↔ if ( 𝑋 ∈ 𝐴 , 1 , 0 ) = 1 ) )
3 eqid ⊢ 1 = 1
4 3 biantru ⊢ ( 𝑋 ∈ 𝐴 ↔ ( 𝑋 ∈ 𝐴 ∧ 1 = 1 ) )
5 ax-1ne0 ⊢ 1 ≠ 0
6 5 neii ⊢ ¬ 1 = 0
7 6 biorfri ⊢ ( ( 𝑋 ∈ 𝐴 ∧ 1 = 1 ) ↔ ( ( 𝑋 ∈ 𝐴 ∧ 1 = 1 ) ∨ 1 = 0 ) )
8 6 bianfi ⊢ ( 1 = 0 ↔ ( ¬ 𝑋 ∈ 𝐴 ∧ 1 = 0 ) )
9 8 orbi2i ⊢ ( ( ( 𝑋 ∈ 𝐴 ∧ 1 = 1 ) ∨ 1 = 0 ) ↔ ( ( 𝑋 ∈ 𝐴 ∧ 1 = 1 ) ∨ ( ¬ 𝑋 ∈ 𝐴 ∧ 1 = 0 ) ) )
10 4 7 9 3bitri ⊢ ( 𝑋 ∈ 𝐴 ↔ ( ( 𝑋 ∈ 𝐴 ∧ 1 = 1 ) ∨ ( ¬ 𝑋 ∈ 𝐴 ∧ 1 = 0 ) ) )
11 eqif ⊢ ( 1 = if ( 𝑋 ∈ 𝐴 , 1 , 0 ) ↔ ( ( 𝑋 ∈ 𝐴 ∧ 1 = 1 ) ∨ ( ¬ 𝑋 ∈ 𝐴 ∧ 1 = 0 ) ) )
12 eqcom ⊢ ( 1 = if ( 𝑋 ∈ 𝐴 , 1 , 0 ) ↔ if ( 𝑋 ∈ 𝐴 , 1 , 0 ) = 1 )
13 10 11 12 3bitr2ri ⊢ ( if ( 𝑋 ∈ 𝐴 , 1 , 0 ) = 1 ↔ 𝑋 ∈ 𝐴 )
14 2 13 bitrdi ⊢ ( ( 𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂 ∧ 𝑋 ∈ 𝑂 ) → ( ( ( ( 𝟭 ‘ 𝑂 ) ‘ 𝐴 ) ‘ 𝑋 ) = 1 ↔ 𝑋 ∈ 𝐴 ) )