Metamath Proof Explorer


Theorem sndisj

Description: Any collection of singletons is disjoint. (Contributed by Mario Carneiro, 14-Nov-2016)

Ref Expression
Assertion sndisj Disj 𝑥 ∈ 𝐴 { 𝑥 }

Proof

Step Hyp Ref Expression
1 dfdisj2 ⊢ ( Disj 𝑥 ∈ 𝐴 { 𝑥 } ↔ ∀ 𝑦 ∃* 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ { 𝑥 } ) )
2 moeq ⊢ ∃* 𝑥 𝑥 = 𝑦
3 simpr ⊢ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ { 𝑥 } ) → 𝑦 ∈ { 𝑥 } )
4 3 elsnd ⊢ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ { 𝑥 } ) → 𝑦 = 𝑥 )
5 4 equcomd ⊢ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ { 𝑥 } ) → 𝑥 = 𝑦 )
6 5 moimi ⊢ ( ∃* 𝑥 𝑥 = 𝑦 → ∃* 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ { 𝑥 } ) )
7 2 6 ax-mp ⊢ ∃* 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ { 𝑥 } )
8 1 7 mpgbir ⊢ Disj 𝑥 ∈ 𝐴 { 𝑥 }