Metamath Proof Explorer


Theorem mopick2

Description: "At most one" can show the existence of a common value. In this case we can infer existence of conjunction from a conjunction of existence, and it is one way to achieve the converse of 19.40 . (Contributed by NM, 5-Apr-2004) (Proof shortened by Andrew Salmon, 9-Jul-2011)

Ref Expression
Assertion mopick2 ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑𝜓 ) ∧ ∃ 𝑥 ( 𝜑𝜒 ) ) → ∃ 𝑥 ( 𝜑𝜓𝜒 ) )

Proof

Step Hyp Ref Expression
1 nfmo1 𝑥 ∃* 𝑥 𝜑
2 nfe1 𝑥𝑥 ( 𝜑𝜓 )
3 1 2 nfan 𝑥 ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑𝜓 ) )
4 mopick ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑𝜓 ) ) → ( 𝜑𝜓 ) )
5 4 ancld ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑𝜓 ) ) → ( 𝜑 → ( 𝜑𝜓 ) ) )
6 5 anim1d ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑𝜓 ) ) → ( ( 𝜑𝜒 ) → ( ( 𝜑𝜓 ) ∧ 𝜒 ) ) )
7 df-3an ( ( 𝜑𝜓𝜒 ) ↔ ( ( 𝜑𝜓 ) ∧ 𝜒 ) )
8 6 7 syl6ibr ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑𝜓 ) ) → ( ( 𝜑𝜒 ) → ( 𝜑𝜓𝜒 ) ) )
9 3 8 eximd ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑𝜓 ) ) → ( ∃ 𝑥 ( 𝜑𝜒 ) → ∃ 𝑥 ( 𝜑𝜓𝜒 ) ) )
10 9 3impia ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑𝜓 ) ∧ ∃ 𝑥 ( 𝜑𝜒 ) ) → ∃ 𝑥 ( 𝜑𝜓𝜒 ) )