zsh substitutions: (P) and ::= operators interaction

Question

Can someone explain to me the result of the following script in zsh:

#!/bin/zsh
var1=var2
var2=first
var3=second
echo var1="$var1"
echo var2="$var2"
echo var3="$var3"
echo '${${(P)var1::=var3}}'="${${(P)var1::=var3}}"
echo var1="$var1"
echo var2="$var2"
echo var3="$var3"
echo ------------
var1=var2
var2=first
var3=second
echo var1="$var1"
echo var2="$var2"
echo var3="$var3"
echo '${(P)var1::=var3}'="${(P)var1::=var3}"
echo var1="$var1"
echo var2="$var2"
echo var3="$var3"

Result:

var1=var2
var2=first
var3=second
${${(P)var1::=var3}}=second
var1=var2
var2=var3
var3=second
------------
var1=var2
var2=first
var3=second
${(P)var1::=var3}=var3
var1=var2
var2=var3
var3=second

As far as I understand the zsh manual (14.3.2 Rules), the execution of the first section (above ------) should be the following:

1. Point 7 (apply the ::= operator) (set var1 to var3)
2. Point 25 (apply the (P) operator) (dereference var1, get var3)
3. Output the value of var3: 'second'

These steps agree with the following line in the result:

${${(P)var1::=var3}}=second

But why in the result the following lines appear:

var1=var2
var2=var3

instead of the expected:

var1=var3
var2=first

The execution of the second section (below ----) should be (according to the manual):

1. Point 4 (apply the (P) operator) (dereference var1, get var2)
2. Point 7 (apply the ::= operator) (set var2 to var3)
3. Output the value of var2: 'var3'

The results of this section match the explanation above perfectly, so the problem is only with the first section.

Answer 1

The (P) applies before the assignment so it's:

1. ${(P)var1::=var3} -> ${var2::=var3}
2. -> ${var3}
3. -> second

with var3 having been assigned to var2.

To force the other order, use ${${(P)${var1::=var3}}}

Note the note in the manual about those rules:

Note that the Zsh Development Group accepts no responsibility for any brain damage which may occur during the reading of the following rules.