What a downcast leaves behind
Cast a wide chain down to u32 and half the assembly simply vanishes. You saw the hint of this on a single op: a u64 + u64 truncated to u32 still emits an addc for a carry nobody reads. Across a chain it goes further. Only the low-word instruction of each step survives; every high-word partner is cut, so a three-operation expression can collapse to two instructions.
Here is low_combo(p, q, r). It subtracts two 64-bit values, ORs in a third, and returns only the low 32 bits:
subfc r4, r6, r4 # (p - q) low word -- carry recorded, never read
or r3, r8, r4 # ( ... | r) low word, delivered straight to r3
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A full-width version would borrow into a subfe and OR the high words as well. Here, nothing of the sort. The u32 return makes every high-word instruction dead code, so MWCC never emits it, and you are left with subfc then or and nothing else. That subfc is the tell. A genuine 32-bit subtract compiles to a bare subf, so the carrying form proves the operands were 64-bit even though only the bottom 32 bits escape to r3.
Your target wears the same shape, a different arithmetic op then a different bitwise op, truncated to u32. Read each surviving low-word instruction back to its C operator, since the cast already pruned the high half and whatever reaches r3 is the whole answer.
Your task
Write and_down_64, taking three u64s and returning a u32, to reproduce the assembly above.