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Theorem issgrpn0 18475
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 18425 . . 3 (𝐴𝐵 → (𝑀 ∈ Mgm ↔ ∀𝑥𝐵𝑦𝐵 (𝑥 𝑦) ∈ 𝐵))
43anbi1d 630 . 2 (𝐴𝐵 → ((𝑀 ∈ Mgm ∧ ∀𝑥𝐵𝑦𝐵𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧))) ↔ (∀𝑥𝐵𝑦𝐵 (𝑥 𝑦) ∈ 𝐵 ∧ ∀𝑥𝐵𝑦𝐵𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧)))))
51, 2issgrp 18473 . 2 (𝑀 ∈ Smgrp ↔ (𝑀 ∈ Mgm ∧ ∀𝑥𝐵𝑦𝐵𝑧𝐵 ((𝑥 𝑦) 𝑧) = (𝑥 (𝑦 𝑧))))
6 r19.26-2 3131 . 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 313 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 205  wa 396   = wceq 1540  wcel 2105  wral 3061  cfv 6479  (class class class)co 7337  Basecbs 17009  +gcplusg 17059  Mgmcmgm 18421  Smgrpcsgrp 18471
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2153  ax-12 2170  ax-ext 2707  ax-nul 5250  ax-pr 5372
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1781  df-nf 1785  df-sb 2067  df-mo 2538  df-eu 2567  df-clab 2714  df-cleq 2728  df-clel 2814  df-ne 2941  df-ral 3062  df-rex 3071  df-rab 3404  df-v 3443  df-sbc 3728  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4270  df-if 4474  df-sn 4574  df-pr 4576  df-op 4580  df-uni 4853  df-br 5093  df-dm 5630  df-iota 6431  df-fv 6487  df-ov 7340  df-mgm 18423  df-sgrp 18472
This theorem is referenced by:  dfgrp3e  18771
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