Theorem irredcl

Description: An irreducible element is in the ring. (Contributed by Mario Carneiro, 4-Dec-2014)

	Ref	Expression
Hypotheses	irredn0.i	⊢ 𝐼 = ( Irred ‘ 𝑅 )
	irredcl.b	⊢ 𝐵 = ( Base ‘ 𝑅 )
Assertion	irredcl	⊢ ( 𝑋 ∈ 𝐼 → 𝑋 ∈ 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		irredn0.i	⊢ 𝐼 = ( Irred ‘ 𝑅 )
2		irredcl.b	⊢ 𝐵 = ( Base ‘ 𝑅 )
3		eqid	⊢ ( Unit ‘ 𝑅 ) = ( Unit ‘ 𝑅 )
4		eqid	⊢ ( .r ‘ 𝑅 ) = ( .r ‘ 𝑅 )
5	2 3 1 4	isirred2	⊢ ( 𝑋 ∈ 𝐼 ↔ ( 𝑋 ∈ 𝐵 ∧ ¬ 𝑋 ∈ ( Unit ‘ 𝑅 ) ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 𝑋 → ( 𝑥 ∈ ( Unit ‘ 𝑅 ) ∨ 𝑦 ∈ ( Unit ‘ 𝑅 ) ) ) ) )
6	5	simp1bi	⊢ ( 𝑋 ∈ 𝐼 → 𝑋 ∈ 𝐵 )