Metamath Proof Explorer

Theorem maxidln1

Description: Obsolete theorem, use mxidln1 instead. One is not contained in any maximal ideal. (Contributed by Jeff Madsen, 17-Jun-2011) (Proof modification is discouraged.) (New usage is discouraged.)

		Ref	Expression
	Hypotheses	maxidln1.1	⊢ 𝐻 = ( 2^nd ‘ 𝑅 )
		maxidln1.2	⊢ 𝑈 = ( GId ‘ 𝐻 )
	Assertion	maxidln1	⊢ ( ( 𝑅 ∈ RingOps ∧ 𝑀 ∈ ( MaxIdl ‘ 𝑅 ) ) → ¬ 𝑈 ∈ 𝑀 )

Proof

Step	Hyp	Ref	Expression
1		maxidln1.1	⊢ 𝐻 = ( 2^nd ‘ 𝑅 )
2		maxidln1.2	⊢ 𝑈 = ( GId ‘ 𝐻 )
3		eqid	⊢ ( 1^st ‘ 𝑅 ) = ( 1^st ‘ 𝑅 )
4		eqid	⊢ ran ( 1^st ‘ 𝑅 ) = ran ( 1^st ‘ 𝑅 )
5	3 4	maxidlnr	⊢ ( ( 𝑅 ∈ RingOps ∧ 𝑀 ∈ ( MaxIdl ‘ 𝑅 ) ) → 𝑀 ≠ ran ( 1^st ‘ 𝑅 ) )
6		maxidlidl	⊢ ( ( 𝑅 ∈ RingOps ∧ 𝑀 ∈ ( MaxIdl ‘ 𝑅 ) ) → 𝑀 ∈ ( Idl ‘ 𝑅 ) )
7	3 1 4 2	1idl	⊢ ( ( 𝑅 ∈ RingOps ∧ 𝑀 ∈ ( Idl ‘ 𝑅 ) ) → ( 𝑈 ∈ 𝑀 ↔ 𝑀 = ran ( 1^st ‘ 𝑅 ) ) )
8	7	necon3bbid	⊢ ( ( 𝑅 ∈ RingOps ∧ 𝑀 ∈ ( Idl ‘ 𝑅 ) ) → ( ¬ 𝑈 ∈ 𝑀 ↔ 𝑀 ≠ ran ( 1^st ‘ 𝑅 ) ) )
9	6 8	syldan	⊢ ( ( 𝑅 ∈ RingOps ∧ 𝑀 ∈ ( MaxIdl ‘ 𝑅 ) ) → ( ¬ 𝑈 ∈ 𝑀 ↔ 𝑀 ≠ ran ( 1^st ‘ 𝑅 ) ) )
10	5 9	mpbird	⊢ ( ( 𝑅 ∈ RingOps ∧ 𝑀 ∈ ( MaxIdl ‘ 𝑅 ) ) → ¬ 𝑈 ∈ 𝑀 )