Proof of Theorem grpoidinvlem2
| Step | Hyp | Ref
| Expression |
| 1 | | simprr 773 |
. . . . 5
⊢ ((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) → 𝐴 ∈ 𝑋) |
| 2 | | simprl 771 |
. . . . 5
⊢ ((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) → 𝑌 ∈ 𝑋) |
| 3 | | grpfo.1 |
. . . . . . . 8
⊢ 𝑋 = ran 𝐺 |
| 4 | 3 | grpocl 30519 |
. . . . . . 7
⊢ ((𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ 𝑌 ∈ 𝑋) → (𝐴𝐺𝑌) ∈ 𝑋) |
| 5 | 4 | 3com23 1127 |
. . . . . 6
⊢ ((𝐺 ∈ GrpOp ∧ 𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋) → (𝐴𝐺𝑌) ∈ 𝑋) |
| 6 | 5 | 3expb 1121 |
. . . . 5
⊢ ((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) → (𝐴𝐺𝑌) ∈ 𝑋) |
| 7 | 1, 2, 6 | 3jca 1129 |
. . . 4
⊢ ((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) → (𝐴 ∈ 𝑋 ∧ 𝑌 ∈ 𝑋 ∧ (𝐴𝐺𝑌) ∈ 𝑋)) |
| 8 | 3 | grpoass 30522 |
. . . 4
⊢ ((𝐺 ∈ GrpOp ∧ (𝐴 ∈ 𝑋 ∧ 𝑌 ∈ 𝑋 ∧ (𝐴𝐺𝑌) ∈ 𝑋)) → ((𝐴𝐺𝑌)𝐺(𝐴𝐺𝑌)) = (𝐴𝐺(𝑌𝐺(𝐴𝐺𝑌)))) |
| 9 | 7, 8 | syldan 591 |
. . 3
⊢ ((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) → ((𝐴𝐺𝑌)𝐺(𝐴𝐺𝑌)) = (𝐴𝐺(𝑌𝐺(𝐴𝐺𝑌)))) |
| 10 | 9 | adantr 480 |
. 2
⊢ (((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) ∧ ((𝑈𝐺𝑌) = 𝑌 ∧ (𝑌𝐺𝐴) = 𝑈)) → ((𝐴𝐺𝑌)𝐺(𝐴𝐺𝑌)) = (𝐴𝐺(𝑌𝐺(𝐴𝐺𝑌)))) |
| 11 | | oveq1 7438 |
. . . . . . 7
⊢ ((𝑌𝐺𝐴) = 𝑈 → ((𝑌𝐺𝐴)𝐺𝑌) = (𝑈𝐺𝑌)) |
| 12 | 11 | adantl 481 |
. . . . . 6
⊢ (((𝑈𝐺𝑌) = 𝑌 ∧ (𝑌𝐺𝐴) = 𝑈) → ((𝑌𝐺𝐴)𝐺𝑌) = (𝑈𝐺𝑌)) |
| 13 | | simpl 482 |
. . . . . 6
⊢ (((𝑈𝐺𝑌) = 𝑌 ∧ (𝑌𝐺𝐴) = 𝑈) → (𝑈𝐺𝑌) = 𝑌) |
| 14 | 12, 13 | eqtr2d 2778 |
. . . . 5
⊢ (((𝑈𝐺𝑌) = 𝑌 ∧ (𝑌𝐺𝐴) = 𝑈) → 𝑌 = ((𝑌𝐺𝐴)𝐺𝑌)) |
| 15 | | id 22 |
. . . . . . 7
⊢ ((𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋 ∧ 𝑌 ∈ 𝑋) → (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋 ∧ 𝑌 ∈ 𝑋)) |
| 16 | 15 | 3anidm13 1422 |
. . . . . 6
⊢ ((𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋) → (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋 ∧ 𝑌 ∈ 𝑋)) |
| 17 | 3 | grpoass 30522 |
. . . . . 6
⊢ ((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋 ∧ 𝑌 ∈ 𝑋)) → ((𝑌𝐺𝐴)𝐺𝑌) = (𝑌𝐺(𝐴𝐺𝑌))) |
| 18 | 16, 17 | sylan2 593 |
. . . . 5
⊢ ((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) → ((𝑌𝐺𝐴)𝐺𝑌) = (𝑌𝐺(𝐴𝐺𝑌))) |
| 19 | 14, 18 | sylan9eqr 2799 |
. . . 4
⊢ (((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) ∧ ((𝑈𝐺𝑌) = 𝑌 ∧ (𝑌𝐺𝐴) = 𝑈)) → 𝑌 = (𝑌𝐺(𝐴𝐺𝑌))) |
| 20 | 19 | eqcomd 2743 |
. . 3
⊢ (((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) ∧ ((𝑈𝐺𝑌) = 𝑌 ∧ (𝑌𝐺𝐴) = 𝑈)) → (𝑌𝐺(𝐴𝐺𝑌)) = 𝑌) |
| 21 | 20 | oveq2d 7447 |
. 2
⊢ (((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) ∧ ((𝑈𝐺𝑌) = 𝑌 ∧ (𝑌𝐺𝐴) = 𝑈)) → (𝐴𝐺(𝑌𝐺(𝐴𝐺𝑌))) = (𝐴𝐺𝑌)) |
| 22 | 10, 21 | eqtrd 2777 |
1
⊢ (((𝐺 ∈ GrpOp ∧ (𝑌 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋)) ∧ ((𝑈𝐺𝑌) = 𝑌 ∧ (𝑌𝐺𝐴) = 𝑈)) → ((𝐴𝐺𝑌)𝐺(𝐴𝐺𝑌)) = (𝐴𝐺𝑌)) |