F' D B' U2 F' B2 L2 U B R U2 R' D2 F2 R' B2 R' B2 U2 B2 (OS)
z2 // Inspection
R // Square
U2 // Pair
D2 B // FB
// Here we see two belt edges in UF and UR with the same side colour, so we know we can solve them together.
E' R2 U' R' // 3QB
u' F' U' F R U' R' // Intuitive EOLE
// There seem to be many unoriented corners, so it would likely be a difficult 6CO case to recognise and execute. So we go ahead with DCAL.
U' R U2 R' U R2 U' R' U2 R' // DCAL
// Since we have CP solved, we can go ahead with CDRLL and get a quick case. Also since the DR edge got solved, we could also go for JTLE which would just be a simple OCLL. In this case, both CDRLL and JTLE have the same algorithm, so it doesn't matter much which reasoning was used.
U R U2 R2 U' R2 U' R2 U2 R // CDRLL / JTLE
M2 U' M' U2 M U' M2 U // Finish
F' D B' U2 F' B2 L2 U B R U2 R' D2 F2 R' B2 R' B2 U2 B2 (TDR)
z2 // Inspection
R // Square
U2 // Pair
D2 B // FB
// Here we see two belt edges in UF and UR with the same side colour, so we know we can solve them together.
E' R2 U' R' // 3QB
u' F' U' F R U' R' // Intuitive EOLE
U' R U2 R' U R2 U' R' U2 R' // TDR
U' R' U R U' R2 U2 R U R' U R2 U' R' U R U' // ZBLL
B2 U2 B D2 B' R2 B' F2 D2 R2 U2 L R D' B R' B2 R' D2 B2 D' (OS)
z' // Inspection
F' L // Square
R' F2 R' U2 F' // FB
// We do not see any obvious ways to solve 2 edges at once, so we proceed to solve the edge that showed up in UF.
U u2 R // 1 edge
// This sets up the UR and DR edges to be belt edges with the same side colour.
E R // 3QB
u' F' U F2 R F' R' // EOLE
// Here, we see a good amount of corners are oriented, so the 6CO case would be easy to recognise and execute.
U2 D R' U R U R' U2 R D' // 6CO
// The DR pieces are all in the U layer, but they're all un-connected, which likely means a bad APDR case. So we simply move forward with 6CP.
U' R2 U R2 U' D R2 U' R2 U R2 D' R2 // 6CP
// Now, we can do the ADF immediately to force an MU-gen L5EP.
D' // ADF
U2 M U2 M' U M' U2 M' U' M2 U2 // L5EP
B2 U2 B D2 B' R2 B' F2 D2 R2 U2 L R D' B R' B2 R' D2 B2 D' (TDR)
z' // Inspection
F' L // Square
R' F2 R' U2 F' // FB
// We do not see any obvious ways to solve 2 edges at once, so we proceed to solve the edge that showed up in UF.
U u2 R // 1 edge
// This sets up the UR and DR edges to be belt edges with the same side colour.
E R // 3QB
u' F' U F2 R F' R' // EOLE
U R2 U R2 F U' R2 U R2 U F' // TDR
u' R U2 R' U' R U' R' U2 R U R' U R U2 R' U2 // ZBLL
// Notice the u' instead of U' for the pre-AUF of ZBLL, to solve the D-offset.