AArch64 also zeroes the upper 32 bits of the destination register when you use a 32 bit instruction.
I know x86-64 zeroes the upper part of the register for backwards compability and improve instruction cache (no need for REX prefix), but AArch64 is unclear for me.
mov w5, w6 // move low 32 bits of register 6 into low 32 bits of register 5
This instruction only depends on the value of register 6. If instead it of zeroing the upper half it left it unchanged, then it would depend on w6 and also the previous value of register 5. That would constrain the renamer and consequently out-of-order execution.