Can a contingent argument be proved invalid with natural deduction?I know that when it is a non-contingent contradiction it can be refuted, but for example in this reasoning:P⊢P∧QI don't know how to refute it.EDIT: What I mean is how someone can figure out that for example the fallacy of Affirming the consequent P→Q,Q⊢P is invalid reasoning without making truth tables. A mathematician could use a contingency in his reasoning inadvertently, so how could he be aware of his error?

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Can a contingent argument be proved invalid with natural deduction?I know that when it is a non-contingent contradiction it can be refuted, but for example in this reasoning:P⊢P∧QI don&#039;t know how to refute it.EDIT: What I mean is how someone can figure out that for example the fallacy of Affirming the consequent P→Q,Q⊢P is invalid reasoning without making truth tables. A mathematician could use a contingency in his reasoning inadvertently, so how could he be aware of his error?

bargeolonakc · Accepted Answer

Natural deduction offers a system for proving sequents, but to show that   Γ  ⊬  ϕ you need an interpretation for the formulae such that   Γ is true (or the conjunction of the formulae in   Γ) and ϕ is false.In your example you can just put P as true and Q as false; since   P  ∧  Q is therefore false we see that   P  ⊬  P  ∧  Q.If you are looking for a more systematic method for finding counterexamples we have truth trees (semantic tableaux).Edit: a proof theoretic approach using cut-free sequent calculus.Suppose   ⇒  P  →    (  P  ∧  Q  ) were derivable with P and Q as atomic formulae.The last rule used must have been R  →, (introduction of   → on the right). That is, we have:            P      ⇒      P      ∧      Q              ⇒      P      →      (      P            ∧            Q      )      And before that we must have had an instance of R  ∧ (introduction of   ∧ on the right) :                                          P            ⇒            P                    ¯                                                        P        ⇒        Q                            P        ⇒        P        ∧        Q                                                                         ⇒        P        →        (        P                ∧                Q        )              P  ⇒  P is axiomatic, which is fine, but there&#039;s no rule with which we can derive Q from P as atomic formulae, so the sequent above is underivable.

Can a contingent argument be proved invalid with natural deduction? I know that when it is a non-contingent contradiction it can be refuted, but for example in this reasoning: PvdashP∧Q

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