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Theorem issgrpn0 18739
Description: The predicate "is a semigroup" for a structure with a nonempty base set. (Contributed by AV, 1-Feb-2020.)
Hypotheses
Ref Expression
issgrpn0.b 𝐵 = (Base‘𝑀)
issgrpn0.o = (+g𝑀)
Assertion
Ref Expression
issgrpn0 (𝐴𝐵 → (𝑀 ∈ Smgrp ↔ ∀𝑥𝐵𝑦𝐵 ((𝑥 𝑦) ∈ 𝐵 ∧ ∀𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧)))))
Distinct variable groups:   𝑥,𝐵,𝑦,𝑧   𝑥,𝑀,𝑦,𝑧   𝑥, ,𝑦,𝑧
Allowed substitution hints:   𝐴(𝑥,𝑦,𝑧)

Proof of Theorem issgrpn0
StepHypRef Expression
1 issgrpn0.b . . . 4 𝐵 = (Base‘𝑀)
2 issgrpn0.o . . . 4 = (+g𝑀)
31, 2ismgmn0 18659 . . 3 (𝐴𝐵 → (𝑀 ∈ Mgm ↔ ∀𝑥𝐵𝑦𝐵 (𝑥 𝑦) ∈ 𝐵))
43anbi1d 640 . 2 (𝐴𝐵 → ((𝑀 ∈ Mgm ∧ ∀𝑥𝐵𝑦𝐵𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧))) ↔ (∀𝑥𝐵𝑦𝐵 (𝑥 𝑦) ∈ 𝐵 ∧ ∀𝑥𝐵𝑦𝐵𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧)))))
51, 2issgrp 18737 . 2 (𝑀 ∈ Smgrp ↔ (𝑀 ∈ Mgm ∧ ∀𝑥𝐵𝑦𝐵𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧))))
6 r19.26-2 3146 . 2 (∀𝑥𝐵𝑦𝐵 ((𝑥 𝑦) ∈ 𝐵 ∧ ∀𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧))) ↔ (∀𝑥𝐵𝑦𝐵 (𝑥 𝑦) ∈ 𝐵 ∧ ∀𝑥𝐵𝑦𝐵𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧))))
74, 5, 63bitr4g 316 1 (𝐴𝐵 → (𝑀 ∈ Smgrp ↔ ∀𝑥𝐵𝑦𝐵 ((𝑥 𝑦) ∈ 𝐵 ∧ ∀𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧)))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 399   = wceq 1559  wcel 2141  wral 3075  cfv 6517  (class class class)co 7392  Basecbs 17228  +gcplusg 17269  Mgmcmgm 18655  Smgrpcsgrp 18735
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733  ax-nul 5255  ax-pr 5389
This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-fal 1572  df-ex 1799  df-sb 2090  df-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-ral 3076  df-rab 3414  df-v 3455  df-sbc 3745  df-dif 3907  df-un 3909  df-ss 3921  df-nul 4286  df-if 4480  df-sn 4582  df-pr 4584  df-op 4588  df-uni 4865  df-br 5100  df-dm 5655  df-iota 6473  df-fv 6525  df-ov 7395  df-mgm 18657  df-sgrp 18736
This theorem is referenced by:  dfgrp3e  19065
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