Metamath Proof Explorer


Theorem disjiunel

Description: A set of elements B of a disjoint set A is disjoint with another element of that set. (Contributed by Thierry Arnoux, 24-May-2020)

Ref Expression
Hypotheses disjiunel.1 ( 𝜑Disj 𝑥𝐴 𝐵 )
disjiunel.2 ( 𝑥 = 𝑌𝐵 = 𝐷 )
disjiunel.3 ( 𝜑𝐸𝐴 )
disjiunel.4 ( 𝜑𝑌 ∈ ( 𝐴𝐸 ) )
Assertion disjiunel ( 𝜑 → ( 𝑥𝐸 𝐵𝐷 ) = ∅ )

Proof

Step Hyp Ref Expression
1 disjiunel.1 ( 𝜑Disj 𝑥𝐴 𝐵 )
2 disjiunel.2 ( 𝑥 = 𝑌𝐵 = 𝐷 )
3 disjiunel.3 ( 𝜑𝐸𝐴 )
4 disjiunel.4 ( 𝜑𝑌 ∈ ( 𝐴𝐸 ) )
5 4 eldifad ( 𝜑𝑌𝐴 )
6 5 snssd ( 𝜑 → { 𝑌 } ⊆ 𝐴 )
7 3 6 unssd ( 𝜑 → ( 𝐸 ∪ { 𝑌 } ) ⊆ 𝐴 )
8 disjss1 ( ( 𝐸 ∪ { 𝑌 } ) ⊆ 𝐴 → ( Disj 𝑥𝐴 𝐵Disj 𝑥 ∈ ( 𝐸 ∪ { 𝑌 } ) 𝐵 ) )
9 7 1 8 sylc ( 𝜑Disj 𝑥 ∈ ( 𝐸 ∪ { 𝑌 } ) 𝐵 )
10 4 eldifbd ( 𝜑 → ¬ 𝑌𝐸 )
11 2 disjunsn ( ( 𝑌𝐴 ∧ ¬ 𝑌𝐸 ) → ( Disj 𝑥 ∈ ( 𝐸 ∪ { 𝑌 } ) 𝐵 ↔ ( Disj 𝑥𝐸 𝐵 ∧ ( 𝑥𝐸 𝐵𝐷 ) = ∅ ) ) )
12 5 10 11 syl2anc ( 𝜑 → ( Disj 𝑥 ∈ ( 𝐸 ∪ { 𝑌 } ) 𝐵 ↔ ( Disj 𝑥𝐸 𝐵 ∧ ( 𝑥𝐸 𝐵𝐷 ) = ∅ ) ) )
13 9 12 mpbid ( 𝜑 → ( Disj 𝑥𝐸 𝐵 ∧ ( 𝑥𝐸 𝐵𝐷 ) = ∅ ) )
14 13 simprd ( 𝜑 → ( 𝑥𝐸 𝐵𝐷 ) = ∅ )