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Theorem mgmidsssn0 18766
Description: Property of the set of identities of 𝐺. Either 𝐺 has no identities, and 𝑂 = ∅, or it has one and this identity is unique and identified by the 0g function. (Contributed by Mario Carneiro, 7-Dec-2014.)
Hypotheses
Ref Expression
mgmidsssn0.b 𝐵 = (Base‘𝐺)
mgmidsssn0.z 0 = (0g𝐺)
mgmidsssn0.p + = (+g𝐺)
mgmidsssn0.o 𝑂 = {𝑥𝐵 ∣ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦)}
Assertion
Ref Expression
mgmidsssn0 (𝐺𝑉𝑂 ⊆ { 0 })
Distinct variable groups:   𝑥,𝑦,𝐵   𝑥,𝐺,𝑦   𝑥, + ,𝑦   𝑥,𝑉   𝑥, 0 ,𝑦
Allowed substitution hints:   𝑂(𝑥, 𝑦)   𝑉(𝑦)

Proof of Theorem mgmidsssn0
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 mgmidsssn0.o . 2 𝑂 = {𝑥𝐵 ∣ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦)}
2 simpr 490 . . . . . . . 8 ((𝐺𝑉 ∧ (𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦))) → (𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦)))
3 mgmidsssn0.b . . . . . . . . 9 𝐵 = (Base‘𝐺)
4 mgmidsssn0.z . . . . . . . . 9 0 = (0g𝐺)
5 mgmidsssn0.p . . . . . . . . 9 + = (+g𝐺)
6 oveq1 7420 . . . . . . . . . . . . 13 (𝑧 = 𝑥 → (𝑧 + 𝑦) = (𝑥 + 𝑦))
76eqeq1d 2762 . . . . . . . . . . . 12 (𝑧 = 𝑥 → ((𝑧 + 𝑦) = 𝑦 ↔ (𝑥 + 𝑦) = 𝑦))
87ovanraleqv 7437 . . . . . . . . . . 11 (𝑧 = 𝑥 → (∀𝑦𝐵 ((𝑧 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑧) = 𝑦) ↔ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦)))
98rspcev 3576 . . . . . . . . . 10 ((𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦)) → ∃𝑧𝐵𝑦𝐵 ((𝑧 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑧) = 𝑦))
109adantl 487 . . . . . . . . 9 ((𝐺𝑉 ∧ (𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦))) → ∃𝑧𝐵𝑦𝐵 ((𝑧 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑧) = 𝑦))
113, 4, 5, 10ismgmid 18758 . . . . . . . 8 ((𝐺𝑉 ∧ (𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦))) → ((𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦)) ↔ 0 = 𝑥))
122, 11mpbid 235 . . . . . . 7 ((𝐺𝑉 ∧ (𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦))) → 0 = 𝑥)
1312eqcomd 2766 . . . . . 6 ((𝐺𝑉 ∧ (𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦))) → 𝑥 = 0 )
14 velsn 4600 . . . . . 6 (𝑥 ∈ { 0 } ↔ 𝑥 = 0 )
1513, 14sylibr 237 . . . . 5 ((𝐺𝑉 ∧ (𝑥𝐵 ∧ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦))) → 𝑥 ∈ { 0 })
1615expr 462 . . . 4 ((𝐺𝑉𝑥𝐵) → (∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦) → 𝑥 ∈ { 0 }))
1716ralrimiva 3154 . . 3 (𝐺𝑉 → ∀𝑥𝐵 (∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦) → 𝑥 ∈ { 0 }))
18 rabss 4018 . . 3 ({𝑥𝐵 ∣ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦)} ⊆ { 0 } ↔ ∀𝑥𝐵 (∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦) → 𝑥 ∈ { 0 }))
1917, 18sylibr 237 . 2 (𝐺𝑉 → {𝑥𝐵 ∣ ∀𝑦𝐵 ((𝑥 + 𝑦) = 𝑦 ∧ (𝑦 + 𝑥) = 𝑦)} ⊆ { 0 })
201, 19eqsstrid 3969 1 (𝐺𝑉𝑂 ⊆ { 0 })
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 401   = wceq 1570  wcel 2145  wral 3076  wrex 3086  {crab 3412  wss 3899  {csn 4584  cfv 6533  (class class class)co 7413  Basecbs 17301  +gcplusg 17342  0gc0g 17524
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 2213  ax-ext 2732  ax-sep 5251  ax-nul 5263  ax-pr 5398
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 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-rex 3087  df-rmo 3365  df-reu 3366  df-rab 3413  df-v 3452  df-sbc 3740  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-mpt 5187  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-iota 6489  df-fun 6535  df-fv 6541  df-riota 7370  df-ov 7416  df-0g 17526
This theorem is used by:  gsumress  18784  gsumval2  18788  gsumvallem2  18943
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