Theorem opcon2b 39657

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 39654	. . . 4 ⊢ ((𝐾 ∈ OP ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘𝑌) ∈ 𝐵)
4	3	3adant2 1132	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘𝑌) ∈ 𝐵)
5	1, 2	opcon3b 39656	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ ( ⊥ ‘𝑌) ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ ( ⊥ ‘( ⊥ ‘𝑌)) = ( ⊥ ‘𝑋)))
6	4, 5	syld3an3 1412	. 2 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ ( ⊥ ‘( ⊥ ‘𝑌)) = ( ⊥ ‘𝑋)))
7	1, 2	opococ 39655	. . . 4 ⊢ ((𝐾 ∈ OP ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘( ⊥ ‘𝑌)) = 𝑌)
8	7	3adant2 1132	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ( ⊥ ‘( ⊥ ‘𝑌)) = 𝑌)
9	8	eqeq1d 2739	. 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 206   ∧ w3a 1087   = wceq 1542   ∈ wcel 2114  ‘cfv 6492  Basecbs 17170  occoc 17219  OPcops 39632

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-nul 5241

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-ral 3053  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-uni 4852  df-br 5087  df-dm 5634  df-iota 6448  df-fv 6500  df-ov 7363  df-oposet 39636

This theorem is referenced by:  opcon1b  39658  riotaocN  39669  glbconxN  39838