Theorem mndissubm 19028

Description: If the base set of a monoid is contained in the base set of another monoid, and the group operation of the monoid is the restriction of the group operation of the other monoid to its base set, and the identity element of the other monoid is contained in the base set of the monoid, then the (base set of the) monoid is a submonoid of the other monoid. Analogous to grpissubg 19383. (Contributed by AV, 17-Feb-2024.)

Hypotheses

Ref	Expression
mndissubm.b	⊢ 𝐵 = (Base‘𝐺)
mndissubm.s	⊢ 𝑆 = (Base‘𝐻)
mndissubm.z	⊢ 0 = (0g‘𝐺)

Assertion

Ref	Expression
mndissubm	⊢ ((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) → ((𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆))) → 𝑆 ∈ (SubMnd‘𝐺)))

Proof of Theorem mndissubm

Dummy variables 𝑎 𝑏 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		simpr1 1213	. . 3 ⊢ (((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) → 𝑆 ⊆ 𝐵)
2		simpr2 1214	. . 3 ⊢ (((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) → 0 ∈ 𝑆)
3		mndmgm 18956	. . . . . . 7 ⊢ (𝐺 ∈ Mnd → 𝐺 ∈ Mgm)
4		mndmgm 18956	. . . . . . 7 ⊢ (𝐻 ∈ Mnd → 𝐻 ∈ Mgm)
5	3, 4	anim12i 625	. . . . . 6 ⊢ ((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) → (𝐺 ∈ Mgm ∧ 𝐻 ∈ Mgm))
6	5	ad2antrr 739	. . . . 5 ⊢ ((((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) ∧ (𝑎 ∈ 𝑆 ∧ 𝑏 ∈ 𝑆)) → (𝐺 ∈ Mgm ∧ 𝐻 ∈ Mgm))
7		3simpb 1167	. . . . . 6 ⊢ ((𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆))) → (𝑆 ⊆ 𝐵 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆))))
8	7	ad2antlr 740	. . . . 5 ⊢ ((((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) ∧ (𝑎 ∈ 𝑆 ∧ 𝑏 ∈ 𝑆)) → (𝑆 ⊆ 𝐵 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆))))
9		simpr 490	. . . . 5 ⊢ ((((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) ∧ (𝑎 ∈ 𝑆 ∧ 𝑏 ∈ 𝑆)) → (𝑎 ∈ 𝑆 ∧ 𝑏 ∈ 𝑆))
10		mndissubm.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐺)
11		mndissubm.s	. . . . . 6 ⊢ 𝑆 = (Base‘𝐻)
12	10, 11	mgmsscl 18847	. . . . 5 ⊢ (((𝐺 ∈ Mgm ∧ 𝐻 ∈ Mgm) ∧ (𝑆 ⊆ 𝐵 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆))) ∧ (𝑎 ∈ 𝑆 ∧ 𝑏 ∈ 𝑆)) → (𝑎(+g‘𝐺)𝑏) ∈ 𝑆)
13	6, 8, 9, 12	syl3anc 1399	. . . 4 ⊢ ((((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) ∧ (𝑎 ∈ 𝑆 ∧ 𝑏 ∈ 𝑆)) → (𝑎(+g‘𝐺)𝑏) ∈ 𝑆)
14	13	ralrimivva 3209	. . 3 ⊢ (((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) → ∀𝑎 ∈ 𝑆 ∀𝑏 ∈ 𝑆 (𝑎(+g‘𝐺)𝑏) ∈ 𝑆)
15		mndissubm.z	. . . . 5 ⊢ 0 = (0g‘𝐺)
16		eqid 2763	. . . . 5 ⊢ (+g‘𝐺) = (+g‘𝐺)
17	10, 15, 16	issubm 19024	. . . 4 ⊢ (𝐺 ∈ Mnd → (𝑆 ∈ (SubMnd‘𝐺) ↔ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ ∀𝑎 ∈ 𝑆 ∀𝑏 ∈ 𝑆 (𝑎(+g‘𝐺)𝑏) ∈ 𝑆)))
18	17	ad2antrr 739	. . 3 ⊢ (((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) → (𝑆 ∈ (SubMnd‘𝐺) ↔ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ ∀𝑎 ∈ 𝑆 ∀𝑏 ∈ 𝑆 (𝑎(+g‘𝐺)𝑏) ∈ 𝑆)))
19	1, 2, 14, 18	mpbir3and 1361	. 2 ⊢ (((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) ∧ (𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆)))) → 𝑆 ∈ (SubMnd‘𝐺))
20	19	ex 418	1 ⊢ ((𝐺 ∈ Mnd ∧ 𝐻 ∈ Mnd) → ((𝑆 ⊆ 𝐵 ∧ 0 ∈ 𝑆 ∧ (+g‘𝐻) = ((+g‘𝐺) ↾ (𝑆 × 𝑆))) → 𝑆 ∈ (SubMnd‘𝐺)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∧ w3a 1103   = wceq 1570   ∈ wcel 2147  ∀wral 3080   ⊆ wss 3900   × cxp 5653   ↾ cres 5657  ‘cfv 6555  (class class class)co 7440  Basecbs 17413  +_gcplusg 17454  0_gc0g 17636  Mgmcmgm 18840  Mndcmnd 18949  SubMndcsubmnd 19003

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5252  ax-nul 5263  ax-pow 5331  ax-pr 5395

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-sbc 3743  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-pw 4562  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-opab 5171  df-mpt 5190  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-res 5667  df-iota 6509  df-fun 6557  df-fv 6563  df-ov 7443  df-mgm 18842  df-sgrp 18934  df-mnd 18950  df-submnd 19005

This theorem is used by:  resmndismnd  19029  submefmnd  19117