Theorem issgrpn0

Description: The predicate "is a semigroup" for a structure with a nonempty base set. (Contributed by AV, 1-Feb-2020)

	Ref	Expression
Hypotheses	issgrpn0.b	⊢ 𝐵 = ( Base ‘ 𝑀 )
	issgrpn0.o	⊢ ⚬ = ( +g ‘ 𝑀 )
Assertion	issgrpn0	⊢ ( 𝐴 ∈ 𝐵 → ( 𝑀 ∈ Smgrp ↔ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ∈ 𝐵 ∧ ∀ 𝑧 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ⚬ 𝑧 ) = ( 𝑥 ⚬ ( 𝑦 ⚬ 𝑧 ) ) ) ) )

Proof

Step	Hyp	Ref	Expression
1		issgrpn0.b	⊢ 𝐵 = ( Base ‘ 𝑀 )
2		issgrpn0.o	⊢ ⚬ = ( +g ‘ 𝑀 )
3	1 2	ismgmn0	⊢ ( 𝐴 ∈ 𝐵 → ( 𝑀 ∈ Mgm ↔ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑥 ⚬ 𝑦 ) ∈ 𝐵 ) )
4	3	anbi1d	⊢ ( 𝐴 ∈ 𝐵 → ( ( 𝑀 ∈ Mgm ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ⚬ 𝑧 ) = ( 𝑥 ⚬ ( 𝑦 ⚬ 𝑧 ) ) ) ↔ ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑥 ⚬ 𝑦 ) ∈ 𝐵 ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ⚬ 𝑧 ) = ( 𝑥 ⚬ ( 𝑦 ⚬ 𝑧 ) ) ) ) )
5	1 2	issgrp	⊢ ( 𝑀 ∈ Smgrp ↔ ( 𝑀 ∈ Mgm ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ⚬ 𝑧 ) = ( 𝑥 ⚬ ( 𝑦 ⚬ 𝑧 ) ) ) )
6		r19.26-2	⊢ ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ∈ 𝐵 ∧ ∀ 𝑧 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ⚬ 𝑧 ) = ( 𝑥 ⚬ ( 𝑦 ⚬ 𝑧 ) ) ) ↔ ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑥 ⚬ 𝑦 ) ∈ 𝐵 ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ⚬ 𝑧 ) = ( 𝑥 ⚬ ( 𝑦 ⚬ 𝑧 ) ) ) )
7	4 5 6	3bitr4g	⊢ ( 𝐴 ∈ 𝐵 → ( 𝑀 ∈ Smgrp ↔ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ∈ 𝐵 ∧ ∀ 𝑧 ∈ 𝐵 ( ( 𝑥 ⚬ 𝑦 ) ⚬ 𝑧 ) = ( 𝑥 ⚬ ( 𝑦 ⚬ 𝑧 ) ) ) ) )