⇤ ← Revision 1 as of 2006-08-22 16:28:38
Size: 512
Comment:
|
Size: 512
Comment:
|
Deletions are marked like this. | Additions are marked like this. |
Line 4: | Line 4: |
Given $(A\wedge B\wedge ...)\vdash C$, and $C\vdash (D\vee E\vee ...)$ we can cut $C$ from the proof of $(D\vee E\vee ...)$ when $(A\wedge B\wedge ...) $ is true. Here we consider $C$ to be the lemma in the prove of $(D\vee E\vee ...)$. | Given $(A\wedge B\wedge ...)\vdash C$, and $C\vdash (D\vee E\vee ...)$ we can cut $C$ from the proof of $(D\vee E\vee ...)$ when $(A\wedge B\wedge ...) $ is true. Here we consider $C$ to be the lemma in the proof of $(D\vee E\vee ...)$. |
Essentially, a cut-free proof is a proof that does not use a lemma. That is
Given $(A\wedge B\wedge ...)\vdash C$, and $C\vdash (D\vee E\vee ...)$ we can cut $C$ from the proof of $(D\vee E\vee ...)$ when $(A\wedge B\wedge ...) $ is true. Here we consider $C$ to be the lemma in the proof of $(D\vee E\vee ...)$. The ability to eliminate $C$ in the above is called the cut-elimination theorem.
For more information set [http://en.wikipedia.org/wiki/Cut-elimination Cut-elimination]