In developing the argument where the premises $p$ and $r$ lead to conclusion $t$, we can use a rule of inference on $p$ such that $p\implies q$ and one on $r$ such that $r\implies s$, after which we can apply another rule of inference on $q$ and $s$ such that $q \wedge s\implies t$, thus showing that $p\wedge r\implies t$.

For example, I was reading the following example:

  1. $\exists x(A(x)\wedge \neg B(x))$ [premise $p$]
  2. $A(c)\wedge \neg B(c)$ [existential instantiation from $p$]
  3. $A(c)$ [simplification from the above; this is statement $q$]
  4. $\forall x(A(x)\rightarrow C(x))$ [premise $r$]
  5. $A(c)\rightarrow C(c)$ [universal instantiation from $r$; this is $s$]
  6. $C(c)$ [modus ponens on $q$ and $s$; this is $t$]

where steps 1 to 3 establish $p\implies q$, steps 4 to 5 establish $r\implies s$, and step 6 is $q\wedge s \implies t$. Intuitively, I can understand why the example works. This feels like a double case of hypothetical syllogism (two chains of implications converging at the modus ponens step), but anyhow I can't justify rigorously through the use of symbols. I've tried $$(p\rightarrow q)\wedge (r \rightarrow s) \wedge (q\wedge s \rightarrow t)\Leftrightarrow(p\wedge r \rightarrow t)$$

but I can't seem to get the manipulations to work. Any guidance will be much appreciated.


The argument needs quantification rules; with only propositional logic we cannot show its validity.

The fact is that there is a "link" between $p \to q$, where $q$ is $A(c)$ and $r \to t$ via $s$, that is $A(c) \to C(c)$.

Without using predicate logic to "see inside" the formulas, the argument :

$(p \to q) \land (r \to s) \vDash (p \land r) \to t$

is not valid.

The argument is :

1) $∃x(A(x) ∧ ¬B(x)) \vdash A(c)$ --- using Existential instantiation

2) $∀x(A(x) → C(x)) \vdash A(c) → C(c)$ --- using Universal instantiation

Thus :

3) $∃x(A(x) ∧ ¬B(x)), ∀x(A(x) → C(x)) \vdash C(c)$ --- by modus ponens.

In conclusion, the issue is not "why can we use multiple rules of inference ?" but "why we have to".

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