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Suppose that I have a proof tree starting with some statement |- B in a sequent calculus, leading to two premises/leaves |- A. Is it always possible to transform such a proof tree into another proof tree, which only contains one leaf |- A only using standard rules of natural deduction?

More schematically, I want to perform the following transformation:

  |- A     |- A                       |- A
------------------    ---->     ---------------
      |- B                            |- B
reinierpost
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aiquita
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2 Answers2

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If I understand your question, I do not think so. That is, maybe the premise $\mathscr{A}$ is used for multiple things in the proof. Consider the following proof of $\mathscr{B}=(A\rightarrow (B\rightarrow C))\rightarrow (A\wedge B \rightarrow C)$:

$$\rightarrow I_2\cfrac{\rightarrow E\cfrac{\rightarrow E \cfrac{\wedge E \cfrac{[ A\wedge B]^1}{A} \quad [A\rightarrow (B\rightarrow C]^2}{B\rightarrow C} \quad \wedge E \cfrac{[A\wedge B]^1}{B}}{\rightarrow I_1\cfrac{C}{A\wedge B \rightarrow C}}}{(A\rightarrow (B\rightarrow C))\rightarrow ((A\wedge B)\rightarrow C)}$$

As you can see in the above proof we have the formula $\mathscr{A} = A\wedge B$ which is used two times. Though both are omitted by using the rule $\rightarrow I_1$, they really need to be used more than once (One time for deduction of $A$ and one time for deduction of $B$).

Doralisa
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"All $A$ are $B$" and "All $B$ are $C$" then "All $A$ are $C$", it is called deductive reasoning.

If two premises are identical then you don't get actually the deduction, you get only the implication.

For example "All $A$ are $B$" then you get the implication "some $A$ are $B.$" "No $A$ are $B$" you get the implication "some $A$ are not $B.$"