Metamath Proof Explorer


Theorem gexid

Description: Any element to the power of the group exponent. (Contributed by Mario Carneiro, 24-Apr-2016)

Ref Expression
Hypotheses gexcl.1 𝑋 = ( Base ‘ 𝐺 )
gexcl.2 𝐸 = ( gEx ‘ 𝐺 )
gexid.3 · = ( .g𝐺 )
gexid.4 0 = ( 0g𝐺 )
Assertion gexid ( 𝐴𝑋 → ( 𝐸 · 𝐴 ) = 0 )

Proof

Step Hyp Ref Expression
1 gexcl.1 𝑋 = ( Base ‘ 𝐺 )
2 gexcl.2 𝐸 = ( gEx ‘ 𝐺 )
3 gexid.3 · = ( .g𝐺 )
4 gexid.4 0 = ( 0g𝐺 )
5 oveq1 ( 𝐸 = 0 → ( 𝐸 · 𝐴 ) = ( 0 · 𝐴 ) )
6 1 4 3 mulg0 ( 𝐴𝑋 → ( 0 · 𝐴 ) = 0 )
7 5 6 sylan9eqr ( ( 𝐴𝑋𝐸 = 0 ) → ( 𝐸 · 𝐴 ) = 0 )
8 7 adantrr ( ( 𝐴𝑋 ∧ ( 𝐸 = 0 ∧ { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } = ∅ ) ) → ( 𝐸 · 𝐴 ) = 0 )
9 oveq1 ( 𝑦 = 𝐸 → ( 𝑦 · 𝑥 ) = ( 𝐸 · 𝑥 ) )
10 9 eqeq1d ( 𝑦 = 𝐸 → ( ( 𝑦 · 𝑥 ) = 0 ↔ ( 𝐸 · 𝑥 ) = 0 ) )
11 10 ralbidv ( 𝑦 = 𝐸 → ( ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 ↔ ∀ 𝑥𝑋 ( 𝐸 · 𝑥 ) = 0 ) )
12 11 elrab ( 𝐸 ∈ { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } ↔ ( 𝐸 ∈ ℕ ∧ ∀ 𝑥𝑋 ( 𝐸 · 𝑥 ) = 0 ) )
13 12 simprbi ( 𝐸 ∈ { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } → ∀ 𝑥𝑋 ( 𝐸 · 𝑥 ) = 0 )
14 oveq2 ( 𝑥 = 𝐴 → ( 𝐸 · 𝑥 ) = ( 𝐸 · 𝐴 ) )
15 14 eqeq1d ( 𝑥 = 𝐴 → ( ( 𝐸 · 𝑥 ) = 0 ↔ ( 𝐸 · 𝐴 ) = 0 ) )
16 15 rspcva ( ( 𝐴𝑋 ∧ ∀ 𝑥𝑋 ( 𝐸 · 𝑥 ) = 0 ) → ( 𝐸 · 𝐴 ) = 0 )
17 13 16 sylan2 ( ( 𝐴𝑋𝐸 ∈ { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } ) → ( 𝐸 · 𝐴 ) = 0 )
18 elfvex ( 𝐴 ∈ ( Base ‘ 𝐺 ) → 𝐺 ∈ V )
19 18 1 eleq2s ( 𝐴𝑋𝐺 ∈ V )
20 eqid { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } = { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 }
21 1 3 4 2 20 gexlem1 ( 𝐺 ∈ V → ( ( 𝐸 = 0 ∧ { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } = ∅ ) ∨ 𝐸 ∈ { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } ) )
22 19 21 syl ( 𝐴𝑋 → ( ( 𝐸 = 0 ∧ { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } = ∅ ) ∨ 𝐸 ∈ { 𝑦 ∈ ℕ ∣ ∀ 𝑥𝑋 ( 𝑦 · 𝑥 ) = 0 } ) )
23 8 17 22 mpjaodan ( 𝐴𝑋 → ( 𝐸 · 𝐴 ) = 0 )