Theorem ltrnatneq

Description: If any atom (under W ) is not equal to its translation, so is any other atom. TODO: -. P .<_ W isn't needed to prove this. Will removing it shorten (and not lengthen) proofs using it? (Contributed by NM, 6-May-2013)

	Ref	Expression
Hypotheses	ltrn2eq.l	⊢ ≤ = ( le ‘ 𝐾 )
	ltrn2eq.a	⊢ 𝐴 = ( Atoms ‘ 𝐾 )
	ltrn2eq.h	⊢ 𝐻 = ( LHyp ‘ 𝐾 )
	ltrn2eq.t	⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
Assertion	ltrnatneq	⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) ∧ ( 𝐹 ‘ 𝑃 ) ≠ 𝑃 ) → ( 𝐹 ‘ 𝑄 ) ≠ 𝑄 )

Proof

Step	Hyp	Ref	Expression
1		ltrn2eq.l	⊢ ≤ = ( le ‘ 𝐾 )
2		ltrn2eq.a	⊢ 𝐴 = ( Atoms ‘ 𝐾 )
3		ltrn2eq.h	⊢ 𝐻 = ( LHyp ‘ 𝐾 )
4		ltrn2eq.t	⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
5	1 2 3 4	ltrn2ateq	⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) ) → ( ( 𝐹 ‘ 𝑃 ) = 𝑃 ↔ ( 𝐹 ‘ 𝑄 ) = 𝑄 ) )
6	5	necon3bid	⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) ) → ( ( 𝐹 ‘ 𝑃 ) ≠ 𝑃 ↔ ( 𝐹 ‘ 𝑄 ) ≠ 𝑄 ) )
7	6	biimp3a	⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝐹 ∈ 𝑇 ∧ ( 𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) ∧ ( 𝐹 ‘ 𝑃 ) ≠ 𝑃 ) → ( 𝐹 ‘ 𝑄 ) ≠ 𝑄 )