Step |
Hyp |
Ref |
Expression |
1 |
|
opoccl.b |
⊢ 𝐵 = ( Base ‘ 𝐾 ) |
2 |
|
opoccl.o |
⊢ ⊥ = ( oc ‘ 𝐾 ) |
3 |
|
fveq2 |
⊢ ( 𝑌 = 𝑋 → ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) ) |
4 |
3
|
eqcoms |
⊢ ( 𝑋 = 𝑌 → ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) ) |
5 |
|
fveq2 |
⊢ ( ( ⊥ ‘ 𝑋 ) = ( ⊥ ‘ 𝑌 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) ) |
6 |
5
|
eqcoms |
⊢ ( ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) ) |
7 |
1 2
|
opococ |
⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = 𝑋 ) |
8 |
7
|
3adant3 |
⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = 𝑋 ) |
9 |
1 2
|
opococ |
⊢ ( ( 𝐾 ∈ OP ∧ 𝑌 ∈ 𝐵 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) = 𝑌 ) |
10 |
9
|
3adant2 |
⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) = 𝑌 ) |
11 |
8 10
|
eqeq12d |
⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ⊥ ‘ ( ⊥ ‘ 𝑋 ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑌 ) ) ↔ 𝑋 = 𝑌 ) ) |
12 |
6 11
|
syl5ib |
⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) → 𝑋 = 𝑌 ) ) |
13 |
4 12
|
impbid2 |
⊢ ( ( 𝐾 ∈ OP ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 = 𝑌 ↔ ( ⊥ ‘ 𝑌 ) = ( ⊥ ‘ 𝑋 ) ) ) |