Metamath Proof Explorer


Theorem eqinfd

Description: Sufficient condition for an element to be equal to the infimum. (Contributed by AV, 3-Sep-2020)

Ref Expression
Hypotheses infexd.1 ⊢ ( 𝜑 → 𝑅 Or 𝐴 )
eqinfd.2 ⊢ ( 𝜑 → 𝐶 ∈ 𝐴 )
eqinfd.3 ⊢ ( ( 𝜑 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝑦 𝑅 𝐶 )
eqinfd.4 ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝐶 𝑅 𝑦 ) ) → ∃ 𝑧 ∈ 𝐵 𝑧 𝑅 𝑦 )
Assertion eqinfd ( 𝜑 → inf ( 𝐵 , 𝐴 , 𝑅 ) = 𝐶 )

Proof

Step Hyp Ref Expression
1 infexd.1 ⊢ ( 𝜑 → 𝑅 Or 𝐴 )
2 eqinfd.2 ⊢ ( 𝜑 → 𝐶 ∈ 𝐴 )
3 eqinfd.3 ⊢ ( ( 𝜑 ∧ 𝑦 ∈ 𝐵 ) → ¬ 𝑦 𝑅 𝐶 )
4 eqinfd.4 ⊢ ( ( 𝜑 ∧ ( 𝑦 ∈ 𝐴 ∧ 𝐶 𝑅 𝑦 ) ) → ∃ 𝑧 ∈ 𝐵 𝑧 𝑅 𝑦 )
5 3 ralrimiva ⊢ ( 𝜑 → ∀ 𝑦 ∈ 𝐵 ¬ 𝑦 𝑅 𝐶 )
6 4 expr ⊢ ( ( 𝜑 ∧ 𝑦 ∈ 𝐴 ) → ( 𝐶 𝑅 𝑦 → ∃ 𝑧 ∈ 𝐵 𝑧 𝑅 𝑦 ) )
7 6 ralrimiva ⊢ ( 𝜑 → ∀ 𝑦 ∈ 𝐴 ( 𝐶 𝑅 𝑦 → ∃ 𝑧 ∈ 𝐵 𝑧 𝑅 𝑦 ) )
8 1 eqinf ⊢ ( 𝜑 → ( ( 𝐶 ∈ 𝐴 ∧ ∀ 𝑦 ∈ 𝐵 ¬ 𝑦 𝑅 𝐶 ∧ ∀ 𝑦 ∈ 𝐴 ( 𝐶 𝑅 𝑦 → ∃ 𝑧 ∈ 𝐵 𝑧 𝑅 𝑦 ) ) → inf ( 𝐵 , 𝐴 , 𝑅 ) = 𝐶 ) )
9 2 5 7 8 mp3and ⊢ ( 𝜑 → inf ( 𝐵 , 𝐴 , 𝑅 ) = 𝐶 )