[PPT] - 63 negations Dave Ripley Universities of Connecticut and Melbourne PowerPoint Presentation

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63 negations

Dave Ripley

Universities of Connecticut and Melbourne

Australasian Association for Logic 2013

davewripley@gmail.com 63 negations

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1024 1024 to 100 100 to 63 63

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1024 DLL, SM 3/ 52

1024

DLL, SM

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(Bounded) DLL: Axioms: A ⊢ A A ⊢ ⊤ ⊥ ⊢ A A0 ∧ A1 ⊢ Ai Ai ⊢ A0 ∨ A1 A ∧ (B ∨ C) ⊢ (A ∧ B) ∨ (A ∧ C) Rules: A ⊢ B A ⊢ C A ⊢ B ∧ C A ⊢ C B ⊢ C A ∨ B ⊢ C A ⊢ B B ⊢ C A ⊢ C

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SM A ⊢ B −B ⊢ −A

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Dualizing a sequent: swap premise/conclusion, ∧/∨, and ⊤/⊥. Every axiom has a dual theorem. Every rule has a dual rule. So every proof has a dual proof: a proof of the dual theorem.

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Derived rule: If A ⊢ B and C( ) is a positive context, then C(A) ⊢ C(B). If C( ) is negative, then C(B) ⊢ C(A). (Proof: induction on C( ).)

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1024 10 principles 8/ 52

1024

10 principles

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Normality principles:

N1: ⊤ ⊢ −⊥ N2: −⊤ ⊢ ⊥ N: N1 + N2

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Antidistribution principles:

A1: −A ∧ −B ⊢ −(A ∨ B) A2: −(A ∧ B) ⊢ −A ∨ −B A: A1 + A2

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Double negation principles:

D1: A ⊢ − − A D2: − − A ⊢ A D: D1 + D2

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Minimal ex’ion principles:

X1: A ∧ −A ⊢ −⊤ X2: −⊥ ⊢ A ∨ −A X: X1 + X2

Recall: −⊤ ⊢ −B, and −B ⊢ −⊥.

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Full ex’ion principles:

X+1: A ∧ −A ⊢ ⊥ X+2: ⊤ ⊢ A ∨ −A X+: X+1 + X+2

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The 1 principles and 2 principles are dual.

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210 = 1024 specifications. How many distinct logics?

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1024 to 100 1024 to 256 16/ 52

1024 to 100

1024 to 256

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Clearly: X1 + N2 entails X+1. Clearly: X+1 entails X1. Less clearly: X+1 entails N2.

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−⊤ ⊢ ⊤ −⊤ ⊢ −⊤

∧R:

−⊤ ⊢ ⊤ ∧ −⊤

X+1:

⊤ ∧ −⊤ ⊢ ⊥

Cut:

−⊤ ⊢ ⊥

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Dually, X+2 entails N1. No need for X+ principles. Down to 28 = 256.

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1024 to 100

256 to 100

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Di entails Ni: ⊥ ⊢ −⊤ − − ⊤ ⊢ −⊥ ⊤ ⊢ − − ⊤ ⊤ ⊢ −⊥

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Di entails Ai:

Part I −A ⊢ −A ∨ −B −(−A ∨ −B) ⊢ − − A

D2:

−(−A ∨ −B) ⊢ A Part II −(−A ∨ −B) ⊢ A ∧ B −(A ∧ B) ⊢ − − (−A ∨ −B)

D2:

−(A ∧ B) ⊢ −A ∨ −B

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NiAi Ni Ai

∅

Di

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This cuts 23 = 8 down to 5. 256 was (8 × 2)2. We reach (5 × 2)2 = 100.

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100 to 63 Remaining entailments 25/ 52

100 to 63

Remaining entailments

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That exhausts one-principle entailments. Combinations remain.

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That exhausts one-principle entailments. Combinations remain. (Note: no use of distribution yet!)

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100 to 63 Remaining entailments 27/ 52

Classical: The pair of X+ principles together entail all others. So X + N does too. Usual presentation of Boolean algebra.

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100 to 63 Remaining entailments 27/ 52

Classical: The pair of X+ principles together entail all others. So X + N does too. Usual presentation of Boolean algebra. Of our 100, 15 are classical.

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100 to 63 Remaining entailments 28/ 52

Full X brings full A. Part I

X2:

−(A ∧ B) ⊢ B ∨ −B

Cut:

−A ∧ A ⊢ −(A ∨ −A)

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100 to 63 Remaining entailments 32/ 52

Xi plus Ni plus Ai brings Ai.

Part I

X1:

−A ∧ − − A ⊲ −(A ∨ B)

Fiddling:

(−A ∧ − − A) ∨ (−A ∧ −(A ∨ B)) ⊲ −(A ∨ B)

Dist:

−A ∧ (− − A ∨ −(A ∨ B)) ⊲ −(A ∨ B)

DRule (A2):

−A ∧ −(−A ∧ (A ∨ B)) ⊲ −(A ∨ B) Part II:

ECQ (X1 + N2):

A ∧ −A ⊲ B

Fiddling:

(−A ∧ A) ∨ (−A ∧ B) ⊲ B

Cut (Dist):

−A ∧ (A ∨ B) ⊲ B

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100 to 63 Remaining entailments 33/ 52

That’s it! Down to 63.

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63 Some interesting critters 34/ 52

63

Some interesting critters

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CL is NX (= DX).

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CL is NX (= DX). FDE is D.

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Preminimal logic (the logic of compatibility frames) is N1A1. Dual premin (the logic of exhaustiveness frames) is N2A2.

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Ockham logic (the logic of the Routley star) is NA.

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D1X1 is the strongest logic here sound for minimal logic, N2D1X1 for intuitionist. D2X2 is the strongest logic here sound for dual minimal logic, N1D2X2 for dual intuitionist.

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D1X1 is the strongest logic here sound for minimal logic, N2D1X1 for intuitionist. D2X2 is the strongest logic here sound for dual minimal logic, N1D2X2 for dual intuitionist.

Are they these logics?

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Other than CL, there are two “attractive” logics among the 100: L1:

X2D1 = XN1 = XA2D1

L2:

X1D2 = XN2 = XA1D2

Anybody recognize these?

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63

Organized by X

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No X: 25 logics X

N1 A1 N1A1 D1

⋆ ⋆ ⋆ ⋆(Pre) ⋆(Quas)

N2

⋆ ⋆ ⋆ ⋆ ⋆

A2

⋆ ⋆ ⋆ ⋆ ⋆

N2A2

⋆(DPre) ⋆ ⋆ ⋆(Ock) ⋆

D2

⋆(DQuas) ⋆ ⋆ ⋆ ⋆(FDE)

63

Organized by N

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XD X1AD1 X2AD2 D X1D1 X2D2 AD1 AD2 X1A1 X2A2 D1 D2 A X1 X2 A1 A2 N

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63 Organized by N 47/ 52

XAD2 X2AD2 X2D2 AD2 X2A D2 X1A X2A2 A X2A1 A2 X1A1 X2 A1 X1 N2

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XAD1 X1AD1 X1D1 AD1 X1A D1 X2A X1A1 A X1A2 A1 X2A2 X1 A2 X2 N1

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∅

X1 X2 A1 A2 X1A2 X2A1 X1A1 X2A2 A X1A X2A XA

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63

Next steps

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Consider the rules of antilogism: Antilogisms A ∧ B ⊢ C A ∧ ¬C ⊢ ¬B A ⊢ B ∨ C ¬B ⊢ ¬A ∨ C A ∧ B ⊢ C ∨ D A ∧ ¬C ⊢ ¬B ∨ D Now how many?

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Which of these critters can coexist?

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Which of these critters can coexist? How do they relate to other critters?

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