Metamath Proof Explorer


Theorem infssd

Description: Inequality deduction for infimum of a subset. (Contributed by AV, 4-Oct-2020)

Ref Expression
Hypotheses infssd.0 ⊢ φ → R Or A
infssd.1 ⊢ φ → C ⊆ B
infssd.3 ⊢ φ → ∃ x ∈ A ∀ y ∈ C ¬ y R x ∧ ∀ y ∈ A x R y → ∃ z ∈ C z R y
infssd.4 ⊢ φ → ∃ x ∈ A ∀ y ∈ B ¬ y R x ∧ ∀ y ∈ A x R y → ∃ z ∈ B z R y
Assertion infssd ⊢ φ → ¬ inf C A R R inf B A R

Proof

Step Hyp Ref Expression
1 infssd.0 ⊢ φ → R Or A
2 infssd.1 ⊢ φ → C ⊆ B
3 infssd.3 ⊢ φ → ∃ x ∈ A ∀ y ∈ C ¬ y R x ∧ ∀ y ∈ A x R y → ∃ z ∈ C z R y
4 infssd.4 ⊢ φ → ∃ x ∈ A ∀ y ∈ B ¬ y R x ∧ ∀ y ∈ A x R y → ∃ z ∈ B z R y
5 1 4 infcl ⊢ φ → inf B A R ∈ A
6 2 sseld ⊢ φ → z ∈ C → z ∈ B
7 1 4 inflb ⊢ φ → z ∈ B → ¬ z R inf B A R
8 6 7 syld ⊢ φ → z ∈ C → ¬ z R inf B A R
9 8 ralrimiv ⊢ φ → ∀ z ∈ C ¬ z R inf B A R
10 1 3 infnlb ⊢ φ → inf B A R ∈ A ∧ ∀ z ∈ C ¬ z R inf B A R → ¬ inf C A R R inf B A R
11 5 9 10 mp2and ⊢ φ → ¬ inf C A R R inf B A R