Metamath Proof Explorer


Theorem invdisj

Description: If there is a function C ( y ) such that C ( y ) = x for all y e. B ( x ) , then the sets B ( x ) for distinct x e. A are disjoint. (Contributed by Mario Carneiro, 10-Dec-2016)

Ref Expression
Assertion invdisj ( ∀ 𝑥𝐴𝑦𝐵 𝐶 = 𝑥Disj 𝑥𝐴 𝐵 )

Proof

Step Hyp Ref Expression
1 nfra2w 𝑦𝑥𝐴𝑦𝐵 𝐶 = 𝑥
2 df-ral ( ∀ 𝑥𝐴𝑦𝐵 𝐶 = 𝑥 ↔ ∀ 𝑥 ( 𝑥𝐴 → ∀ 𝑦𝐵 𝐶 = 𝑥 ) )
3 rsp ( ∀ 𝑦𝐵 𝐶 = 𝑥 → ( 𝑦𝐵𝐶 = 𝑥 ) )
4 eqcom ( 𝐶 = 𝑥𝑥 = 𝐶 )
5 3 4 syl6ib ( ∀ 𝑦𝐵 𝐶 = 𝑥 → ( 𝑦𝐵𝑥 = 𝐶 ) )
6 5 imim2i ( ( 𝑥𝐴 → ∀ 𝑦𝐵 𝐶 = 𝑥 ) → ( 𝑥𝐴 → ( 𝑦𝐵𝑥 = 𝐶 ) ) )
7 6 impd ( ( 𝑥𝐴 → ∀ 𝑦𝐵 𝐶 = 𝑥 ) → ( ( 𝑥𝐴𝑦𝐵 ) → 𝑥 = 𝐶 ) )
8 7 alimi ( ∀ 𝑥 ( 𝑥𝐴 → ∀ 𝑦𝐵 𝐶 = 𝑥 ) → ∀ 𝑥 ( ( 𝑥𝐴𝑦𝐵 ) → 𝑥 = 𝐶 ) )
9 2 8 sylbi ( ∀ 𝑥𝐴𝑦𝐵 𝐶 = 𝑥 → ∀ 𝑥 ( ( 𝑥𝐴𝑦𝐵 ) → 𝑥 = 𝐶 ) )
10 mo2icl ( ∀ 𝑥 ( ( 𝑥𝐴𝑦𝐵 ) → 𝑥 = 𝐶 ) → ∃* 𝑥 ( 𝑥𝐴𝑦𝐵 ) )
11 9 10 syl ( ∀ 𝑥𝐴𝑦𝐵 𝐶 = 𝑥 → ∃* 𝑥 ( 𝑥𝐴𝑦𝐵 ) )
12 1 11 alrimi ( ∀ 𝑥𝐴𝑦𝐵 𝐶 = 𝑥 → ∀ 𝑦 ∃* 𝑥 ( 𝑥𝐴𝑦𝐵 ) )
13 dfdisj2 ( Disj 𝑥𝐴 𝐵 ↔ ∀ 𝑦 ∃* 𝑥 ( 𝑥𝐴𝑦𝐵 ) )
14 12 13 sylibr ( ∀ 𝑥𝐴𝑦𝐵 𝐶 = 𝑥Disj 𝑥𝐴 𝐵 )