Metamath Proof Explorer


Theorem difprsnss

Description: Removal of a singleton from an unordered pair. (Contributed by NM, 16-Mar-2006) (Proof shortened by Andrew Salmon, 29-Jun-2011)

Ref Expression
Assertion difprsnss ⊢ A B ∖ A ⊆ B

Proof

Step Hyp Ref Expression
1 vex ⊢ x ∈ V
2 1 elpr ⊢ x ∈ A B ↔ x = A ∨ x = B
3 velsn ⊢ x ∈ A ↔ x = A
4 3 notbii ⊢ ¬ x ∈ A ↔ ¬ x = A
5 biorf ⊢ ¬ x = A → x = B ↔ x = A ∨ x = B
6 5 biimparc ⊢ x = A ∨ x = B ∧ ¬ x = A → x = B
7 2 4 6 syl2anb ⊢ x ∈ A B ∧ ¬ x ∈ A → x = B
8 eldif ⊢ x ∈ A B ∖ A ↔ x ∈ A B ∧ ¬ x ∈ A
9 velsn ⊢ x ∈ B ↔ x = B
10 7 8 9 3imtr4i ⊢ x ∈ A B ∖ A → x ∈ B
11 10 ssriv ⊢ A B ∖ A ⊆ B