Metamath Proof Explorer


Theorem ress0gOLD

Description: Obsolete version of ress0g as of 12-Aug-2026. 0g is unaffected by restriction. This is a bit more generic than submnd0 . (Contributed by Thierry Arnoux, 23-Oct-2017) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypotheses ress0g.s ⊢ S = R ↾ 𝑠 A
ress0g.b ⊢ B = Base R
ress0g.0 ⊢ 0 ˙ = 0 R
Assertion ress0gOLD ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B → 0 ˙ = 0 S

Proof

Step Hyp Ref Expression
1 ress0g.s ⊢ S = R ↾ 𝑠 A
2 ress0g.b ⊢ B = Base R
3 ress0g.0 ⊢ 0 ˙ = 0 R
4 1 2 ressbas2 ⊢ A ⊆ B → A = Base S
5 4 3ad2ant3 ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B → A = Base S
6 simp3 ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B → A ⊆ B
7 2 fvexi ⊢ B ∈ V
8 ssexg ⊢ A ⊆ B ∧ B ∈ V → A ∈ V
9 6 7 8 sylancl ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B → A ∈ V
10 eqid ⊢ + R = + R
11 1 10 ressplusg ⊢ A ∈ V → + R = + S
12 9 11 syl ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B → + R = + S
13 simp2 ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B → 0 ˙ ∈ A
14 simpl1 ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B ∧ x ∈ A → R ∈ Mnd
15 6 sselda ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B ∧ x ∈ A → x ∈ B
16 2 10 3 mndlid ⊢ R ∈ Mnd ∧ x ∈ B → 0 ˙ + R x = x
17 14 15 16 syl2anc ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B ∧ x ∈ A → 0 ˙ + R x = x
18 2 10 3 mndrid ⊢ R ∈ Mnd ∧ x ∈ B → x + R 0 ˙ = x
19 14 15 18 syl2anc ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B ∧ x ∈ A → x + R 0 ˙ = x
20 5 12 13 17 19 grpidd ⊢ R ∈ Mnd ∧ 0 ˙ ∈ A ∧ A ⊆ B → 0 ˙ = 0 S