Metamath Proof Explorer


Theorem nelsubgcld

Description: A non-subgroup-member plus a subgroup member is a non-subgroup-member. (Contributed by Steven Nguyen, 15-Apr-2023)

Ref Expression
Hypotheses nelsubginvcld.g ⊢ ( 𝜑 → 𝐺 ∈ Grp )
nelsubginvcld.s ⊢ ( 𝜑 → 𝑆 ∈ ( SubGrp ‘ 𝐺 ) )
nelsubginvcld.x ⊢ ( 𝜑 → 𝑋 ∈ ( 𝐵 ∖ 𝑆 ) )
nelsubginvcld.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
nelsubgcld.y ⊢ ( 𝜑 → 𝑌 ∈ 𝑆 )
nelsubgcld.p ⊢ + = ( +g ‘ 𝐺 )
Assertion nelsubgcld ( 𝜑 → ( 𝑋 + 𝑌 ) ∈ ( 𝐵 ∖ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 nelsubginvcld.g ⊢ ( 𝜑 → 𝐺 ∈ Grp )
2 nelsubginvcld.s ⊢ ( 𝜑 → 𝑆 ∈ ( SubGrp ‘ 𝐺 ) )
3 nelsubginvcld.x ⊢ ( 𝜑 → 𝑋 ∈ ( 𝐵 ∖ 𝑆 ) )
4 nelsubginvcld.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
5 nelsubgcld.y ⊢ ( 𝜑 → 𝑌 ∈ 𝑆 )
6 nelsubgcld.p ⊢ + = ( +g ‘ 𝐺 )
7 3 eldifad ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
8 4 subgss ⊢ ( 𝑆 ∈ ( SubGrp ‘ 𝐺 ) → 𝑆 ⊆ 𝐵 )
9 2 8 syl ⊢ ( 𝜑 → 𝑆 ⊆ 𝐵 )
10 9 5 sseldd ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
11 4 6 grpcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + 𝑌 ) ∈ 𝐵 )
12 1 7 10 11 syl3anc ⊢ ( 𝜑 → ( 𝑋 + 𝑌 ) ∈ 𝐵 )
13 3 eldifbd ⊢ ( 𝜑 → ¬ 𝑋 ∈ 𝑆 )
14 eqid ⊢ ( -g ‘ 𝐺 ) = ( -g ‘ 𝐺 )
15 4 6 14 grppncan ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + 𝑌 ) ( -g ‘ 𝐺 ) 𝑌 ) = 𝑋 )
16 1 7 10 15 syl3anc ⊢ ( 𝜑 → ( ( 𝑋 + 𝑌 ) ( -g ‘ 𝐺 ) 𝑌 ) = 𝑋 )
17 16 adantr ⊢ ( ( 𝜑 ∧ ( 𝑋 + 𝑌 ) ∈ 𝑆 ) → ( ( 𝑋 + 𝑌 ) ( -g ‘ 𝐺 ) 𝑌 ) = 𝑋 )
18 2 adantr ⊢ ( ( 𝜑 ∧ ( 𝑋 + 𝑌 ) ∈ 𝑆 ) → 𝑆 ∈ ( SubGrp ‘ 𝐺 ) )
19 simpr ⊢ ( ( 𝜑 ∧ ( 𝑋 + 𝑌 ) ∈ 𝑆 ) → ( 𝑋 + 𝑌 ) ∈ 𝑆 )
20 5 adantr ⊢ ( ( 𝜑 ∧ ( 𝑋 + 𝑌 ) ∈ 𝑆 ) → 𝑌 ∈ 𝑆 )
21 14 subgsubcl ⊢ ( ( 𝑆 ∈ ( SubGrp ‘ 𝐺 ) ∧ ( 𝑋 + 𝑌 ) ∈ 𝑆 ∧ 𝑌 ∈ 𝑆 ) → ( ( 𝑋 + 𝑌 ) ( -g ‘ 𝐺 ) 𝑌 ) ∈ 𝑆 )
22 18 19 20 21 syl3anc ⊢ ( ( 𝜑 ∧ ( 𝑋 + 𝑌 ) ∈ 𝑆 ) → ( ( 𝑋 + 𝑌 ) ( -g ‘ 𝐺 ) 𝑌 ) ∈ 𝑆 )
23 17 22 eqeltrrd ⊢ ( ( 𝜑 ∧ ( 𝑋 + 𝑌 ) ∈ 𝑆 ) → 𝑋 ∈ 𝑆 )
24 13 23 mtand ⊢ ( 𝜑 → ¬ ( 𝑋 + 𝑌 ) ∈ 𝑆 )
25 12 24 eldifd ⊢ ( 𝜑 → ( 𝑋 + 𝑌 ) ∈ ( 𝐵 ∖ 𝑆 ) )