Theorem opcon1b 40071

Description: Orthocomplement contraposition law. (negcon1 11534 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 40070	. . 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 1570   ∈ wcel 2145  ‘cfv 6533  Basecbs 17301  occoc 17350  OPcops 40045

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732  ax-nul 5263

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-dm 5665  df-iota 6489  df-fv 6541  df-ov 7416  df-oposet 40049

This theorem is used by:  opoc0  40076