Theorem opcon2b 39689

Description: Orthocomplement contraposition law. (negcon2 11438 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 39686	. . . 4 ⊢ ((𝐾 ∈ OP ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘𝑌) ∈ 𝐵)
4	3	3adant2 1137	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘𝑌) ∈ 𝐵)
5	1, 2	opcon3b 39688	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ ( ⊥ ‘𝑌) ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ ( ⊥ ‘( ⊥ ‘𝑌)) = ( ⊥ ‘𝑋)))
6	4, 5	syld3an3 1417	. 2 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ ( ⊥ ‘( ⊥ ‘𝑌)) = ( ⊥ ‘𝑋)))
7	1, 2	opococ 39687	. . . 4 ⊢ ((𝐾 ∈ OP ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘( ⊥ ‘𝑌)) = 𝑌)
8	7	3adant2 1137	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘( ⊥ ‘𝑌)) = 𝑌)
9	8	eqeq1d 2741	. 2 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (( ⊥ ‘( ⊥ ‘𝑌)) = ( ⊥ ‘𝑋) ↔ 𝑌 = ( ⊥ ‘𝑋)))
10	6, 9	bitrd 280	1 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ 𝑌 = ( ⊥ ‘𝑋)))

Syntax hints:   → wi 4   ↔ wb 207   ∧ w3a 1092   = wceq 1547   ∈ wcel 2119  ‘cfv 6485  Basecbs 17170  occoc 17219  OPcops 39664

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711  ax-nul 5228

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-ne 2935  df-ral 3054  df-rab 3392  df-v 3433  df-dif 3886  df-un 3888  df-ss 3900  df-nul 4262  df-if 4455  df-sn 4556  df-pr 4558  df-op 4562  df-uni 4839  df-br 5073  df-dm 5628  df-iota 6441  df-fv 6493  df-ov 7359  df-oposet 39668

This theorem is referenced by:  opcon1b  39690  riotaocN  39701  glbconxN  39870