Theorem opcon1b 40349

Description: Orthocomplement contraposition law. (negcon1 11614 analog.) (Contributed by NM, 24-Jan-2012.)

Hypotheses

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

Assertion

Ref	Expression
opcon1b	⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (( ⊥ ‘𝑋) = 𝑌 ↔ ( ⊥ ‘𝑌) = 𝑋))

Proof of Theorem opcon1b

Step	Hyp	Ref	Expression
1		opoccl.b	. . . 4 ⊢ 𝐵 = (Base‘𝐾)
2		opoccl.o	. . . 4 ⊢ ⊥ = (oc‘𝐾)
3	1, 2	opcon2b 40348	. . 3 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 = ( ⊥ ‘𝑌) ↔ 𝑌 = ( ⊥ ‘𝑋)))
4		eqcom 2767	. . 3 ⊢ (( ⊥ ‘𝑌) = 𝑋 ↔ 𝑋 = ( ⊥ ‘𝑌))
5		eqcom 2767	. . 3 ⊢ (( ⊥ ‘𝑋) = 𝑌 ↔ 𝑌 = ( ⊥ ‘𝑋))
6	3, 4, 5	3bitr4g 317	. 2 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (( ⊥ ‘𝑌) = 𝑋 ↔ ( ⊥ ‘𝑋) = 𝑌))
7	6	bicomd 226	1 ⊢ ((𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (( ⊥ ‘𝑋) = 𝑌 ↔ ( ⊥ ‘𝑌) = 𝑋))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ w3a 1103   = wceq 1569   ∈ wcel 2144  ‘cfv 6539  Basecbs 17393  occoc 17442  OPcops 40323

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-nul 5259

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-un 4151  df-nul 4279  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-uni 4867  df-br 5103  df-dm 5660  df-iota 6494  df-fv 6547  df-ov 7424  df-oposet 40327

This theorem is used by:  opoc0  40354