Metamath Proof Explorer


Theorem ress0g

Description: 0g is unaffected by restriction. This is a bit more generic than submnd0 . (Contributed by Thierry Arnoux, 23-Oct-2017) (Proof shortened by AV, 12-Aug-2026)

Ref Expression
Hypotheses ress0g.s 𝑆 = ( 𝑅s 𝐴 )
ress0g.b 𝐵 = ( Base ‘ 𝑅 )
ress0g.0 0 = ( 0g𝑅 )
Assertion ress0g ( ( 𝑅 ∈ Mnd ∧ 0𝐴𝐴𝐵 ) → 0 = ( 0g𝑆 ) )

Proof

Step Hyp Ref Expression
1 ress0g.s 𝑆 = ( 𝑅s 𝐴 )
2 ress0g.b 𝐵 = ( Base ‘ 𝑅 )
3 ress0g.0 0 = ( 0g𝑅 )
4 eqid ( +g𝑅 ) = ( +g𝑅 )
5 2 4 mndid ( 𝑅 ∈ Mnd → ∃ 𝑢𝐵𝑥𝐵 ( ( 𝑢 ( +g𝑅 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( +g𝑅 ) 𝑢 ) = 𝑥 ) )
6 5 3ad2ant1 ( ( 𝑅 ∈ Mnd ∧ 0𝐴𝐴𝐵 ) → ∃ 𝑢𝐵𝑥𝐵 ( ( 𝑢 ( +g𝑅 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( +g𝑅 ) 𝑢 ) = 𝑥 ) )
7 simp3 ( ( 𝑅 ∈ Mnd ∧ 0𝐴𝐴𝐵 ) → 𝐴𝐵 )
8 simp2 ( ( 𝑅 ∈ Mnd ∧ 0𝐴𝐴𝐵 ) → 0𝐴 )
9 2 4 3 6 1 7 8 idressid ( ( 𝑅 ∈ Mnd ∧ 0𝐴𝐴𝐵 ) → ( 0g𝑆 ) = 0 )
10 9 eqcomd ( ( 𝑅 ∈ Mnd ∧ 0𝐴𝐴𝐵 ) → 0 = ( 0g𝑆 ) )