Metamath Proof Explorer


Theorem oppgmndb

Description: Bidirectional form of oppgmnd . (Contributed by Stefan O'Rear, 26-Aug-2015)

Ref Expression
Hypothesis oppgbas.1 ⊢ O = opp 𝑔 ⁡ R
Assertion oppgmndb ⊢ R ∈ Mnd ↔ O ∈ Mnd

Proof

Step Hyp Ref Expression
1 oppgbas.1 ⊢ O = opp 𝑔 ⁡ R
2 1 oppgmnd ⊢ R ∈ Mnd → O ∈ Mnd
3 eqid ⊢ opp 𝑔 ⁡ O = opp 𝑔 ⁡ O
4 3 oppgmnd ⊢ O ∈ Mnd → opp 𝑔 ⁡ O ∈ Mnd
5 eqid ⊢ Base R = Base R
6 1 5 oppgbas ⊢ Base R = Base O
7 3 6 oppgbas ⊢ Base R = Base opp 𝑔 ⁡ O
8 7 a1i ⊢ ⊤ → Base R = Base opp 𝑔 ⁡ O
9 eqidd ⊢ ⊤ → Base R = Base R
10 eqid ⊢ + O = + O
11 eqid ⊢ + opp 𝑔 ⁡ O = + opp 𝑔 ⁡ O
12 10 3 11 oppgplus ⊢ x + opp 𝑔 ⁡ O y = y + O x
13 eqid ⊢ + R = + R
14 13 1 10 oppgplus ⊢ y + O x = x + R y
15 12 14 eqtri ⊢ x + opp 𝑔 ⁡ O y = x + R y
16 15 a1i ⊢ ⊤ ∧ x ∈ Base R ∧ y ∈ Base R → x + opp 𝑔 ⁡ O y = x + R y
17 8 9 16 mndpropd ⊢ ⊤ → opp 𝑔 ⁡ O ∈ Mnd ↔ R ∈ Mnd
18 17 mptru ⊢ opp 𝑔 ⁡ O ∈ Mnd ↔ R ∈ Mnd
19 4 18 sylib ⊢ O ∈ Mnd → R ∈ Mnd
20 2 19 impbii ⊢ R ∈ Mnd ↔ O ∈ Mnd