About an article by Jeroen Valk

Key reference is: Propositional Paradoxes. At Logical Preliminaries we have the following table:

PP1 $(p \wedge \neg p) \rightarrow q$ from contradiction everything follows

PP2 $p \rightarrow (q \vee \neg q)$ everything implies a logical tautology

PP3 $\neg p \rightarrow (p \rightarrow q)$ from false propositions everything follows

PP4 $q \rightarrow (p \rightarrow q)$ everything implies a true proposition

PP5 $(p \rightarrow q) \vee (q \rightarrow r)$ imply or be implied!

PP6 $[ (p \rightarrow q) \wedge (r \rightarrow s) ] \rightarrow [ (p \rightarrow s) \vee (r \rightarrow q) ]$ consequent exchange

PP7 $[ (p \wedge q) \rightarrow r ] \rightarrow [ (p \rightarrow r) \vee (q \rightarrow r) ]$ Rutten's paradox

As has been explained extensively at the Mathematics Stack Exchange forum, these Propositional Paradoxes shall be analyzed according to the sequential interpretation.
As usual, 1 stands for TRUE and 0 stands for FALSE. Don't cares are indicated by X. Here goes!

PP1: from contradiction everything follows

(if p then (if p then 0 else 1) else 0) then q else 1 : general
(if 0                           else 0) [    ] else 1 : p=0,q=X
(if 1 then (if 1 then 0       )         [    ] else 1 : p=1,q=X

It is clear that $q$ is an unemployed variable. Hence, in the sequential interpretation of propositional logic, q is simply not present, as indicated by [].
According to our hypothesis, this explains the apparent paradox in logica naturalis.

PP2: everything implies a logical tautology

if p then (if q then 1 else (if q then 0 else 1)) else 1 : general
if 0                                              else 1 : p=0,q=X
if 1 then (if 0        else (if 0        else 1))        : p=1,q=0
if 1 then (if 1 then 1                          )        : p=1,q=1

It is clear that all instances of the logical variables are functional in the sequential interpretation.
Hence, according to our hypothesis, there is no reason to consider PP2 as a paradox in logica naturalis.

PP3: from false propositions everything follows

if (if p then 0 else 1) then (if p then q else 1) else 1 : general
if (if 0        else 1) then (if 0 [    ] else 1)        : p=0,q=X
if (if 1 then 0       )            [    ]         else 1 : p=1,q=X

It is clear that $q$ is an unemployed variable. Hence, in the sequential interpretation of propositional logic, q is simply not present.
According to our hypothesis, this explains the apparent paradox in logica naturalis.

PP4: everything implies a true proposition

if q then (if p then q else 1) else 1 : general
if 0                           else 1 : p=X,q=0
if 1 then (if 0        else 1)        : p=0,q=1
if 1 then (if 1 then 1       )        : p=1,q=1

It is clear that all instances of the logical variables are functional in the sequential interpretation.
Hence, according to our hypothesis, there is no reason to consider PP4 as a paradox in logica naturalis.

PP5: imply or be implied!

if (if p then q else 1) then 1 else (if q then r else 1) : general
if (if 0        else 1) then 1            [    ]         : p=0,q=X,r=X
if (if 1 then 0       )        else (if 0 [    ] else 1) : p=1,q=0,r=X
if (if 1 then 1       ) then 1            [    ]         : p=1,q=1,r=X

It is clear that $r$ is an unemployed variable. Hence, in the sequential interpretation of propositional logic, r is simply not present.
According to our hypothesis, this explains the apparent paradox in logica naturalis.

PP6: consequent exchange

We split in two halves, as follows:

if [ first half ] then [ second half ] else 1

If the [ first half ] is 0 (FALSE) then the program jumps to the 'else', giving 1 (TRUE). So, for the [ second half ], we only have to consider [ first half ] outcomes with (1).
First half:

if (if p then q else 1) then (if r then s else 1) else 0 : general
if (if 0        else 1) then (if 0        else 1)        : p=0,q=X,r=0,s=X (1)
if (if 0        else 1) then (if 1 then 0       )        : p=0,q=X,r=1,s=0 (0)
if (if 0        else 1) then (if 1 then 1       )        : p=0,q=X,r=1,s=1 (1)
if (if 1 then 0       )                           else 0 : p=1,q=0,r=X,s=X (0)
if (if 1 then 1       ) then (if 0        else 1)        : p=1,q=1,r=0,s=X (1)
if (if 1 then 1       ) then (if 1 then 0       )        : p=1,q=1,r=1,s=0 (0)
if (if 1 then 1       ) then (if 1 then 1       )        : p=1,q=1,r=1,s=1 (1)

Second half:

if (if p then s else 1) then 1 else (if r then q else 1) : general
if (if 0        else 1) then 1            [    ]         : p=0,q=X,r=X,s=X
if (if 1 then 0       )        else (if 0 [    ] else 1) : p=1,q=X,r=0,s=0
if (if 1 then 1       ) then 1            [    ]         : p=1,q=X,r=0,s=1

It is clear that the last instance of $q$ is unemployed. Hence, in the sequential interpretation of propositional logic, q is simply not present there [].
According to our hypothesis, this explains the apparent paradox in logica naturalis.

PP7: Rutten's paradox

is at the Mathematics Stack Exchange forum. We are finished.

According to our hypothesis, PP1, PP3, PP5, PP6, PP7 are paradoxes in logica naturalis while PP2 and PP4 are not.

PP1	$(p \wedge \neg p) \rightarrow q$	from contradiction everything follows
PP2	$p \rightarrow (q \vee \neg q)$	everything implies a logical tautology
PP3	$\neg p \rightarrow (p \rightarrow q)$	from false propositions everything follows
PP4	$q \rightarrow (p \rightarrow q)$	everything implies a true proposition
PP5	$(p \rightarrow q) \vee (q \rightarrow r)$	imply or be implied!
PP6	$[ (p \rightarrow q) \wedge (r \rightarrow s) ] \rightarrow [ (p \rightarrow s) \vee (r \rightarrow q) ]$	consequent exchange
PP7	$[ (p \wedge q) \rightarrow r ] \rightarrow [ (p \rightarrow r) \vee (q \rightarrow r) ]$	Rutten's paradox