Theorem issgrpALT

Description: The predicate "is a semigroup". (Contributed by AV, 16-Jan-2020) (New usage is discouraged.) (Proof modification is discouraged.)

	Ref	Expression
Hypotheses	ismgmALT.b	$$⊢B={\mathrm{Base}}_M$$
	ismgmALT.o	No typesetting found for |- .o. = ( +g ` M ) with typecode |-
Assertion	issgrpALT	Could not format assertion : No typesetting found for |- ( M e. SGrpALT <-> ( M e. MgmALT /\ .o. assLaw B ) ) with typecode |-

Proof

Step	Hyp	Ref	Expression
1		ismgmALT.b	$$⊢B={\mathrm{Base}}_M$$
2		ismgmALT.o	Could not format .o. = ( +g ` M ) : No typesetting found for |- .o. = ( +g ` M ) with typecode |-
3		fveq2	$$⊢m=M\to {+}_m={+}_M$$
4	3 2	eqtr4di	Could not format ( m = M -> ( +g ` m ) = .o. ) : No typesetting found for |- ( m = M -> ( +g ` m ) = .o. ) with typecode |-
5		fveq2	$$⊢m=M\to {\mathrm{Base}}_m={\mathrm{Base}}_M$$
6	5 1	eqtr4di	$$⊢m=M\to {\mathrm{Base}}_m=B$$
7	4 6	breq12d	Could not format ( m = M -> ( ( +g ` m ) assLaw ( Base ` m ) <-> .o. assLaw B ) ) : No typesetting found for |- ( m = M -> ( ( +g ` m ) assLaw ( Base ` m ) <-> .o. assLaw B ) ) with typecode |-
8		df-sgrp2	$$⊢\mathrm{SGrpALT}=\left\{m\in \mathrm{MgmALT}|{+}_m\mathrm{assLaw}{\mathrm{Base}}_m\right\}$$
9	7 8	elrab2	Could not format ( M e. SGrpALT <-> ( M e. MgmALT /\ .o. assLaw B ) ) : No typesetting found for |- ( M e. SGrpALT <-> ( M e. MgmALT /\ .o. assLaw B ) ) with typecode |-