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