Metamath Proof Explorer


Theorem plydivlem1

Description: Lemma for plydivalg . (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Hypotheses plydiv.pl ⊢ φ ∧ x ∈ S ∧ y ∈ S → x + y ∈ S
plydiv.tm ⊢ φ ∧ x ∈ S ∧ y ∈ S → x ⁢ y ∈ S
plydiv.rc ⊢ φ ∧ x ∈ S ∧ x ≠ 0 → 1 x ∈ S
plydiv.m1 ⊢ φ → − 1 ∈ S
Assertion plydivlem1 ⊢ φ → 0 ∈ S

Proof

Step Hyp Ref Expression
1 plydiv.pl ⊢ φ ∧ x ∈ S ∧ y ∈ S → x + y ∈ S
2 plydiv.tm ⊢ φ ∧ x ∈ S ∧ y ∈ S → x ⁢ y ∈ S
3 plydiv.rc ⊢ φ ∧ x ∈ S ∧ x ≠ 0 → 1 x ∈ S
4 plydiv.m1 ⊢ φ → − 1 ∈ S
5 1pneg1e0 ⊢ 1 + -1 = 0
6 neg1mulneg1e1 ⊢ -1 ⁢ -1 = 1
7 2 4 4 caovcld ⊢ φ → -1 ⁢ -1 ∈ S
8 6 7 eqeltrrid ⊢ φ → 1 ∈ S
9 1 8 4 caovcld ⊢ φ → 1 + -1 ∈ S
10 5 9 eqeltrrid ⊢ φ → 0 ∈ S