Theorem opcon2b 38579

Description: Orthocomplement contraposition law. (negcon2 11514 analog.) (Contributed by NM, 16-Jan-2012.)

Hypotheses

Ref	Expression
opoccl.b	⊢ 𝐵 = (Base‘𝐾)
opoccl.o	⊢ ⊥ = (oc‘𝐾)

Assertion

Ref	Expression
opcon2b	⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ 𝑌 = ( ⊥ ‘𝑋)))

Proof of Theorem opcon2b

Step	Hyp	Ref	Expression
1		opoccl.b	. . . . 5 ⊢ 𝐵 = (Base‘𝐾)
2		opoccl.o	. . . . 5 ⊢ ⊥ = (oc‘𝐾)
3	1, 2	opoccl 38576	. . . 4 ⊢ ((𝐾 ∈ OP ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘𝑌) ∈ 𝐵)
4	3	3adant2 1128	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘𝑌) ∈ 𝐵)
5	1, 2	opcon3b 38578	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ ( ⊥ ‘𝑌) ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ ( ⊥ ‘( ⊥ ‘𝑌)) = ( ⊥ ‘𝑋)))
6	4, 5	syld3an3 1406	. 2 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ ( ⊥ ‘( ⊥ ‘𝑌)) = ( ⊥ ‘𝑋)))
7	1, 2	opococ 38577	. . . 4 ⊢ ((𝐾 ∈ OP ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘( ⊥ ‘𝑌)) = 𝑌)
8	7	3adant2 1128	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘( ⊥ ‘𝑌)) = 𝑌)
9	8	eqeq1d 2728	. 2 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (( ⊥ ‘( ⊥ ‘𝑌)) = ( ⊥ ‘𝑋) ↔ 𝑌 = ( ⊥ ‘𝑋)))
10	6, 9	bitrd 279	1 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ 𝑌 = ( ⊥ ‘𝑋)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ w3a 1084   = wceq 1533   ∈ wcel 2098  ‘cfv 6536  Basecbs 17150  occoc 17211  OPcops 38554

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2697  ax-nul 5299

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-ne 2935  df-ral 3056  df-rab 3427  df-v 3470  df-dif 3946  df-un 3948  df-in 3950  df-ss 3960  df-nul 4318  df-if 4524  df-sn 4624  df-pr 4626  df-op 4630  df-uni 4903  df-br 5142  df-dm 5679  df-iota 6488  df-fv 6544  df-ov 7407  df-oposet 38558

This theorem is referenced by:  opcon1b  38580  riotaocN  38591  glbconxN  38761