But with left shift available we can now also implement right shift because for a 32 bit register, right shift by N positions can be interpreted as a left shift by 32 - N positions and than looking at the upper 32 bits. This is visualized below

The verilog code needs to be changed only a little bit:

```
wire shiftq = op[4:0] == 12; // true if operaration is shift left
wire shiftqr = op[4:0] == 13; // true if operaration is shift right
wire doshift = shiftq | shiftqr;
wire [5:0] invertshift = 6'd32 - {1'b0,b[4:0]};
wire [4:0] nshift = shiftqr ? invertshift[4:0] : b[4:0];
wire shiftlo = doshift & ~nshift[4]; // true if shifting < 16 bits
wire shifthi = doshift & nshift[4]; // true if shifting >= 16 bits
...
// 4 16x16 bit partial multiplications
// the multiplier is either the b operand or a power of two for a shift
// note that b[31:16] for shift operations [31-0] is always zero
// so when shiftlo is true al_bh and ah_bh still result in zero
// the same is not true the other way around hence the extra shiftq check
// note that the behavior is undefined for shifts > 31
wire [31:0] mult_al_bl = a[15: 0] * (shiftlo ? shiftla16 : doshift ? 16'b0 : b[15: 0]);
wire [31:0] mult_al_bh = a[15: 0] * (shifthi ? shiftla16 : b[31:16]);
wire [31:0] mult_ah_bl = a[31:16] * (shiftlo ? shiftla16 : doshift ? 16'b0 : b[15: 0]);
wire [31:0] mult_ah_bh = a[31:16] * (shifthi ? shiftla16 : b[31:16]);
...
assign result =
...
shiftq ? {1'b0, mult64[31:0]} :
shiftqr ? {1'b0, mult64[63:32]} :
...
;
```

The only thing we do here is subtracting the number of positions to shift from 32 if we are dealing with a shift right instruction and also swap in the correct arguments for the multiplication for both the left and the right shift operation. Also, when selecting the final result we take care of selecting the uppermost 32 bits fro the right shift where a left shift would select the lower 32 bits.