Theorem mndissubm 18916

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 19271. (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 18845	. . . . . . 7 ⊢ (𝐺 ∈ Mnd → 𝐺 ∈ Mgm)
4		mndmgm 18845	. . . . . . 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 18739	. . . . 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 1398	. . . 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 3207	. . 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 2762	. . . . 5 ⊢ (+g‘𝐺) = (+g‘𝐺)
17	10, 15, 16	issubm 18912	. . . 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 2145  ∀wral 3078   ⊆ wss 3902   × cxp 5657   ↾ cres 5661  ‘cfv 6537  (class class class)co 7416  Basecbs 17305  +_gcplusg 17346  0_gc0g 17528  Mgmcmgm 18732  Mndcmnd 18838  SubMndcsubmnd 18891

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734  ax-sep 5255  ax-nul 5267  ax-pow 5334  ax-pr 5402

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 2100  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-sbc 3743  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-pw 4562  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-opab 5172  df-mpt 5191  df-id 5554  df-xp 5665  df-rel 5666  df-cnv 5667  df-co 5668  df-dm 5669  df-res 5671  df-iota 6493  df-fun 6539  df-fv 6545  df-ov 7419  df-mgm 18734  df-sgrp 18823  df-mnd 18839  df-submnd 18893

This theorem is used by:  resmndismnd  18917  submefmnd  19005