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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 ( 𝜑 ∧ 𝜓 ) ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜒 ) ) → ∃ 𝑥 ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 nfmo1 ⊢ Ⅎ 𝑥 ∃* 𝑥 𝜑
2 nfe1 ⊢ Ⅎ 𝑥 ∃ 𝑥 ( 𝜑 ∧ 𝜓 )
3 1 2 nfan ⊢ Ⅎ 𝑥 ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) )
4 mopick ⊢ ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ) → ( 𝜑 → 𝜓 ) )
5 4 ancld ⊢ ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ) → ( 𝜑 → ( 𝜑 ∧ 𝜓 ) ) )
6 5 anim1d ⊢ ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ) → ( ( 𝜑 ∧ 𝜒 ) → ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) ) )
7 df-3an ⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) )
8 6 7 imbitrrdi ⊢ ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ) → ( ( 𝜑 ∧ 𝜒 ) → ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ) )
9 3 8 eximd ⊢ ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ) → ( ∃ 𝑥 ( 𝜑 ∧ 𝜒 ) → ∃ 𝑥 ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ) )
10 9 3impia ⊢ ( ( ∃* 𝑥 𝜑 ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜓 ) ∧ ∃ 𝑥 ( 𝜑 ∧ 𝜒 ) ) → ∃ 𝑥 ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) )