Rule of Replacement and Rule of Inference My question is how I can solve this argument. Can you please help me?


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*$(V\implies \lnot W)\land(X\implies Y)$

*$(\lnot W\implies Z)\land(Y\implies\lnot A)$

*$(Z\implies\lnot B)\land(\lnot A\implies C)$

*$V\land X\therefore \lnot B\land C$
 A: Each step you will need to prove this uses the same laws of deduction: hypothetical syllogism and $\land$-elimination, except that at the end you need $\land$-introduction. Or maybe you're using a whole different sort of system. You havem't told us.
A: I don't know what your "primitive" or "basic" or "non-derived" rules of inference are.  So, I'll derive your conclusion using just four rules and the assumptions you have given (otherwise I'd need axioms... and your system might not have any axioms... heck it might not even qualify as a system at all since it might not have any well-formed formulas, like my first symbolic logic text).


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*Detachment (abbreviated Co): From ($\alpha\implies\beta$) as well as $\alpha$, we may infer $\beta$

*Conjunction-in (abbreviated Ki): From $\alpha$, as well as $\beta$ we may infer ($\alpha \land \beta$)

*Conjunction-out left (abbreviated Ko-l): From ($\alpha\land\beta$), we may infer $\alpha$.

*Conjunction-out right (abbreviated Ko-r): From ($\alpha\land\beta$), we may infer $\beta$.


You can "solve" this by discharging all the conjunctions with the conjunction-out rules, then using the detachment rule several times, and then for the very last step use the conjunction-in rule.  Your text may call those rules names like "simplification, modus ponens, and something else" or "conjunction elimination, modus ponens, and conjunction introduction."  If you have a rule which says something like "from $(\alpha \implies \beta)$, as well as $(\beta \implies\gamma)$, we may infer $(\alpha\implies\gamma)$,"  you can solve this in another way, but then you'll probably end up using more rules of inference, so I prefer not to do that here.  But, you might want to write that proof, since it comes as shorter in length (though I suspect many logicians probably would maintain it as taking more for granted than what I do below on very solid grounds).
I strongly suggest that you try writing the proof now before reading the following or only read parts of the proof and try to finish it yourself as you read.
Your first assumption goes


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*$(V\implies \lnot W)\land(X\implies Y)$ your second assumption goes

*$(\lnot W\implies Z)\land(Y\implies \lnot A)$ your third assumption goes

*$(Z\implies\lnot B)\land(\lnot A\implies C)$ your fourth assumption goes

*$(V\land X)$   by Ko-l on 4 we get 5

*$V$    by Ko-r on 4 we get 6

*$X$    by Ko-l on 1 we get 7

*$(V\implies \lnot W)$ by Ko-r on 1 we get 8

*$(X\implies Y)$ by Ko-l on 2 we get 9

*$(\lnot W\implies Z)$ by Ko-r on 2 we get 10

*$(Y\implies \lnot A)$ by Ko-l on 3 we get 11

*$(Z\implies \lnot B)$ by Ko-r on 3 we get 12

*$(\lnot A\implies C)$ by 7, 5 Co we get 13

*$\lnot W$ by 8, 6 Co we get 14

*$Y$   by 13, 9 Co we get 15

*$Z$   by 14, 10 Co we get 16

*$\lnot A$  by 15, 11 Co we get 17

*$\lnot B$  by 16, 12 Co we get 18

*$C$  by 17, 18 Ki we get 19

*$(\lnot B \land C)$
A: $\fbox{Given}$
[1] $(V\implies \lnot W)\land(X\implies Y)$
[2] $(\lnot W\implies Z)\land(Y\implies\lnot A)$
[3] $(Z\implies\lnot B)\land(\lnot A\implies C)$
$\fbox{Assume}$ ($V\land X$). From ($V\land X$) by conjunction elimination:
[4] $V$
[5] $X$
From (1–3) by the transitivity of the arrow:
[6] $V \Longrightarrow \lnot B$
[7] $X \Longrightarrow \lnot C$
From (4 & 6) and (5 & 7) by modus ponens: 
[8] $\lnot B$
[9] $C$
Conjoining (8 & 9):
[10] $~\lnot B \land C$
$\fbox{Therefore}$ ($1$–$3$), ($V \land X$) $\vdash$ ($\lnot B \land C$)
A: this is a formal proof that needs to be completed.
one possible completion of the proof is
5. v          step 4, simplification
6. v>~w       1 simplification
7. ~w>z       2 simplification
8. z>~B       3 simplification
9. ~w         5,6 modus ponens
10. z         9,7 MP
11. ~b        8,10 MP
12. x>y       1 simplification
13. y>~a      2 simplification
14. ~a>c      3 simplification
15. x         4 simplification
16. y         12,15 MP
17. ~a        13,16 MP
18. c         14,17 MP
therefore ~b^c          11,18 addition/conjunction
most of this proof is just modus ponens and simplification. if you are at least somewhat familiar with these laws of inference, this proof should make sense now
