Metamath Proof Explorer


Theorem plydivlem1

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

Ref Expression
Hypotheses plydiv.pl ( ( 𝜑 ∧ ( 𝑥𝑆𝑦𝑆 ) ) → ( 𝑥 + 𝑦 ) ∈ 𝑆 )
plydiv.tm ( ( 𝜑 ∧ ( 𝑥𝑆𝑦𝑆 ) ) → ( 𝑥 · 𝑦 ) ∈ 𝑆 )
plydiv.rc ( ( 𝜑 ∧ ( 𝑥𝑆𝑥 ≠ 0 ) ) → ( 1 / 𝑥 ) ∈ 𝑆 )
plydiv.m1 ( 𝜑 → - 1 ∈ 𝑆 )
Assertion plydivlem1 ( 𝜑 → 0 ∈ 𝑆 )

Proof

Step Hyp Ref Expression
1 plydiv.pl ( ( 𝜑 ∧ ( 𝑥𝑆𝑦𝑆 ) ) → ( 𝑥 + 𝑦 ) ∈ 𝑆 )
2 plydiv.tm ( ( 𝜑 ∧ ( 𝑥𝑆𝑦𝑆 ) ) → ( 𝑥 · 𝑦 ) ∈ 𝑆 )
3 plydiv.rc ( ( 𝜑 ∧ ( 𝑥𝑆𝑥 ≠ 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 )