Metamath Proof Explorer


Theorem orngmul

Description: In an ordered ring, the ordering is compatible with the ring multiplication operation. (Contributed by Thierry Arnoux, 20-Jan-2018)

Ref Expression
Hypotheses orngmul.0 𝐵 = ( Base ‘ 𝑅 )
orngmul.1 = ( le ‘ 𝑅 )
orngmul.2 0 = ( 0g𝑅 )
orngmul.3 · = ( .r𝑅 )
Assertion orngmul ( ( 𝑅 ∈ oRing ∧ ( 𝑋𝐵0 𝑋 ) ∧ ( 𝑌𝐵0 𝑌 ) ) → 0 ( 𝑋 · 𝑌 ) )

Proof

Step Hyp Ref Expression
1 orngmul.0 𝐵 = ( Base ‘ 𝑅 )
2 orngmul.1 = ( le ‘ 𝑅 )
3 orngmul.2 0 = ( 0g𝑅 )
4 orngmul.3 · = ( .r𝑅 )
5 simp2r ( ( 𝑅 ∈ oRing ∧ ( 𝑋𝐵0 𝑋 ) ∧ ( 𝑌𝐵0 𝑌 ) ) → 0 𝑋 )
6 simp3r ( ( 𝑅 ∈ oRing ∧ ( 𝑋𝐵0 𝑋 ) ∧ ( 𝑌𝐵0 𝑌 ) ) → 0 𝑌 )
7 simp2l ( ( 𝑅 ∈ oRing ∧ ( 𝑋𝐵0 𝑋 ) ∧ ( 𝑌𝐵0 𝑌 ) ) → 𝑋𝐵 )
8 simp3l ( ( 𝑅 ∈ oRing ∧ ( 𝑋𝐵0 𝑋 ) ∧ ( 𝑌𝐵0 𝑌 ) ) → 𝑌𝐵 )
9 1 3 4 2 isorng ( 𝑅 ∈ oRing ↔ ( 𝑅 ∈ Ring ∧ 𝑅 ∈ oGrp ∧ ∀ 𝑎𝐵𝑏𝐵 ( ( 0 𝑎0 𝑏 ) → 0 ( 𝑎 · 𝑏 ) ) ) )
10 9 simp3bi ( 𝑅 ∈ oRing → ∀ 𝑎𝐵𝑏𝐵 ( ( 0 𝑎0 𝑏 ) → 0 ( 𝑎 · 𝑏 ) ) )
11 10 3ad2ant1 ( ( 𝑅 ∈ oRing ∧ ( 𝑋𝐵0 𝑋 ) ∧ ( 𝑌𝐵0 𝑌 ) ) → ∀ 𝑎𝐵𝑏𝐵 ( ( 0 𝑎0 𝑏 ) → 0 ( 𝑎 · 𝑏 ) ) )
12 breq2 ( 𝑎 = 𝑋 → ( 0 𝑎0 𝑋 ) )
13 12 anbi1d ( 𝑎 = 𝑋 → ( ( 0 𝑎0 𝑏 ) ↔ ( 0 𝑋0 𝑏 ) ) )
14 oveq1 ( 𝑎 = 𝑋 → ( 𝑎 · 𝑏 ) = ( 𝑋 · 𝑏 ) )
15 14 breq2d ( 𝑎 = 𝑋 → ( 0 ( 𝑎 · 𝑏 ) ↔ 0 ( 𝑋 · 𝑏 ) ) )
16 13 15 imbi12d ( 𝑎 = 𝑋 → ( ( ( 0 𝑎0 𝑏 ) → 0 ( 𝑎 · 𝑏 ) ) ↔ ( ( 0 𝑋0 𝑏 ) → 0 ( 𝑋 · 𝑏 ) ) ) )
17 breq2 ( 𝑏 = 𝑌 → ( 0 𝑏0 𝑌 ) )
18 17 anbi2d ( 𝑏 = 𝑌 → ( ( 0 𝑋0 𝑏 ) ↔ ( 0 𝑋0 𝑌 ) ) )
19 oveq2 ( 𝑏 = 𝑌 → ( 𝑋 · 𝑏 ) = ( 𝑋 · 𝑌 ) )
20 19 breq2d ( 𝑏 = 𝑌 → ( 0 ( 𝑋 · 𝑏 ) ↔ 0 ( 𝑋 · 𝑌 ) ) )
21 18 20 imbi12d ( 𝑏 = 𝑌 → ( ( ( 0 𝑋0 𝑏 ) → 0 ( 𝑋 · 𝑏 ) ) ↔ ( ( 0 𝑋0 𝑌 ) → 0 ( 𝑋 · 𝑌 ) ) ) )
22 16 21 rspc2va ( ( ( 𝑋𝐵𝑌𝐵 ) ∧ ∀ 𝑎𝐵𝑏𝐵 ( ( 0 𝑎0 𝑏 ) → 0 ( 𝑎 · 𝑏 ) ) ) → ( ( 0 𝑋0 𝑌 ) → 0 ( 𝑋 · 𝑌 ) ) )
23 7 8 11 22 syl21anc ( ( 𝑅 ∈ oRing ∧ ( 𝑋𝐵0 𝑋 ) ∧ ( 𝑌𝐵0 𝑌 ) ) → ( ( 0 𝑋0 𝑌 ) → 0 ( 𝑋 · 𝑌 ) ) )
24 5 6 23 mp2and ( ( 𝑅 ∈ oRing ∧ ( 𝑋𝐵0 𝑋 ) ∧ ( 𝑌𝐵0 𝑌 ) ) → 0 ( 𝑋 · 𝑌 ) )