Chapter 13
Cut Elimination (Hauptsatz)
Gentzen's theorem, one of the most important in logic, is not very far removed from normalisation in natural deduction, which is to a large extent inspired by it. In a slightly modi ed form, it is at the root of languages such as PROLOG. In other words, it is a result which everyone should see proved at least once. However the proof is very delicate and ddly. So we shall begin by pointing out the key cases which it is important to understand. Afterwards we shall develop the detailed proof, whose intricacies are less interesting.
13.1The key cases
The aim is to eliminate cuts of the special form
A ` C; B A0; C ` B0
Cut
A; A0 ` B; B0
where the left premise is a right logical rule and the right premise a left logical rule, so that both introduce the main symbol of C. These cases enlighten the deep symmetries of logical rules, which match each other exactly.
1. R^ and L1^ |
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A ` C; B A0 ` D; B0 |
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A00; C ` B00 |
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` C ^ D; B; B |
0 |
R^ |
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; C ^ D ` B |
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L1^ |
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A; A |
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A |
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Cut |
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A; A0; A00 ` B; B0; B00 |
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is replaced by
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13.1. THE KEY CASES |
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A ` C; B A00; C ` B00
Cut
A; A00 ` B; B00
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A; A0; A00 ` B; B0; B00
where the double bar denotes a certain number of structural rules, in this case weakening and exchange.
2. R^ and L2^ |
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A ` C; B A0 ` D; B0 |
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A00; D ` B00 |
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` C ^ D; B; B |
0 |
R^ |
00 |
; C ^ D ` B |
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L2^ |
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A; A |
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A |
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Cut |
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A; A0; A00 ` B; B0; B00 |
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is replaced similarly by
A0 ` D; B0 A00; D ` B00
Cut
A0; A00 ` B0; B00
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A; A0; A00 ` B; B0; B00
3. R1_ and L_ |
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A ` C; B |
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A0; C ` B0 |
A00; D ` B00 |
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R1_ |
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L_ |
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A ` C _ D; B |
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A |
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_ D ` B |
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Cut |
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A; A0; A00 ` B; B0; B00 |
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is replaced by
A ` C; B A0; C ` B0
Cut
A; A0 ` B; B0
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A; A0; A00 ` B; B0; B00
This is the dual of case 1.
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CHAPTER 13. CUT ELIMINATION (HAUPTSATZ) |
4. R2_ and L_ |
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A ` D; B |
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A0; C ` B0 |
A00; D ` B00 |
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R2_ |
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L_ |
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A ` C _ D; B |
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A |
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_ D ` B |
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Cut |
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A; A0; A00 ` B; B0; B00 |
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is replaced by
A ` D; B A00; D ` B00
Cut
A; A00 ` B; B00
================
A; A0; A00 ` B; B0; B00
This is the dual of case 2.
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A; C ` B |
A0 ` C; B0 |
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R: |
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:C ` B |
0 |
L: |
5. R: and L: |
A ` :C; B |
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Cut |
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A; A0 ` B; B0 |
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is replaced by
A0 ` C; B0 A; C ` B
Cut
A0; A ` B0; B
==========
A; A0 ` B; B0
Note the switch. |
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6. R) and L) |
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A; C ` D; B |
A0 ` C; B0 |
A00; D ` B00 |
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R) |
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L) |
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A ` C ) D; B |
A |
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; C ) D ` B |
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Cut |
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A; A0; A00 ` B; B0; B00 |
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is replaced by
13.1. THE KEY CASES |
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A0 ` C; B0 |
A; C ` D; B |
Cut |
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A0; A ` B0; D; B |
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A00; D ` B00 |
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A; A0 |
` D; B; B0 |
Cut |
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A; A0; A00 ` B; B0; B00 |
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So the problem is solved by two cuts.
7. R8 and L8
A ` C; B |
A0; C[a= ] ` B0 |
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R8 |
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; 8 : C ` B |
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L8 |
A ` 8 : C; B |
A |
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Cut |
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A; A0 ` B; B0 |
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is replaced by
A ` C[a= ]; B A0; C[a= ] ` B0
Cut
A; A0 ` B; B0
where a is substituted for throughout the left-hand sub-proof.
8. R9 and L9
A ` C[a= ]; B |
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A0; C ` B0 |
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R9 |
0 |
; 9 : C ` B |
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L9 |
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A ` 9 : C; B |
A |
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Cut |
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A; A0 ` B; B0 |
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is replaced by
A ` C[a= ]; B A0; C[a= ] ` B0
Cut
A; A0 ` B; B0
This is the dual of case 7.