Metamath Proof Explorer


Theorem issubgrpd

Description: Prove a subgroup by closure. (Contributed by Stefan O'Rear, 7-Dec-2014)

Ref Expression
Hypotheses issubgrpd.s ( 𝜑𝑆 = ( 𝐼s 𝐷 ) )
issubgrpd.z ( 𝜑0 = ( 0g𝐼 ) )
issubgrpd.p ( 𝜑+ = ( +g𝐼 ) )
issubgrpd.ss ( 𝜑𝐷 ⊆ ( Base ‘ 𝐼 ) )
issubgrpd.zcl ( 𝜑0𝐷 )
issubgrpd.acl ( ( 𝜑𝑥𝐷𝑦𝐷 ) → ( 𝑥 + 𝑦 ) ∈ 𝐷 )
issubgrpd.ncl ( ( 𝜑𝑥𝐷 ) → ( ( invg𝐼 ) ‘ 𝑥 ) ∈ 𝐷 )
issubgrpd.g ( 𝜑𝐼 ∈ Grp )
Assertion issubgrpd ( 𝜑𝑆 ∈ Grp )

Proof

Step Hyp Ref Expression
1 issubgrpd.s ( 𝜑𝑆 = ( 𝐼s 𝐷 ) )
2 issubgrpd.z ( 𝜑0 = ( 0g𝐼 ) )
3 issubgrpd.p ( 𝜑+ = ( +g𝐼 ) )
4 issubgrpd.ss ( 𝜑𝐷 ⊆ ( Base ‘ 𝐼 ) )
5 issubgrpd.zcl ( 𝜑0𝐷 )
6 issubgrpd.acl ( ( 𝜑𝑥𝐷𝑦𝐷 ) → ( 𝑥 + 𝑦 ) ∈ 𝐷 )
7 issubgrpd.ncl ( ( 𝜑𝑥𝐷 ) → ( ( invg𝐼 ) ‘ 𝑥 ) ∈ 𝐷 )
8 issubgrpd.g ( 𝜑𝐼 ∈ Grp )
9 1 2 3 4 5 6 7 8 issubgrpd2 ( 𝜑𝐷 ∈ ( SubGrp ‘ 𝐼 ) )
10 eqid ( 𝐼s 𝐷 ) = ( 𝐼s 𝐷 )
11 10 subggrp ( 𝐷 ∈ ( SubGrp ‘ 𝐼 ) → ( 𝐼s 𝐷 ) ∈ Grp )
12 9 11 syl ( 𝜑 → ( 𝐼s 𝐷 ) ∈ Grp )
13 1 12 eqeltrd ( 𝜑𝑆 ∈ Grp )