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Decidability of logical theories

CSC 4170 Theory of Computation. Decidability of logical theories. Section 6.2. 6.2.a. Giorgi Japaridze Theory of Computability. Formulas --- strings produced by the following CFG:. The language of arithmetic: formulas. FORMULA  ATOM | ( FORMULA ) |

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Decidability of logical theories

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  1. CSC 4170 Theory of Computation Decidability of logical theories Section 6.2

  2. 6.2.a Giorgi JaparidzeTheory of Computability Formulas --- strings produced by the following CFG: The language of arithmetic: formulas FORMULA  ATOM | (FORMULA) | (FORMULA)  (FORMULA) | (FORMULA)  (FORMULA) |  VARIABLE (FORMULA) ATOM  TERM = TERM TERM  VARIABLE | CONSTANT | (TERM) + (TERM) (TERM)  (TERM) VARIABLE  v | VARIABLE’ CONSTANT  0 | 1 | 1CONSTANT negation conjunction disjunction universal quantifier

  3. 6.2.b Giorgi JaparidzeTheory of Computability An occurrence of variable x is bound in formula F, if it is in the scope of x, i.e. F= ... x( … x …) … Otherwise it is free. The language of arithmetic: sentences A sentence is a formula without free occurrences of variables Is v’ free or bound in: v’=0  (v’(v’=0))  (v’’(v’=0)) v’’(v’=v’’) v’((v’=0) (v’=v)) (v’(v’=0))(v’=v) Is the following formula a sentence: (0=10) v=0 v(v=v) v(v=v’) v(v=1 v’(v+v’=110))

  4. 6.2.c Giorgi JaparidzeTheory of Computability Truth of arithmetic sentences • An atomic sentence is true iff it is true under the standard • interpretation of constants and +,,=. •  A is true iff A is false • AB is true iff both A and B are true • AB is true iff either A or B (or both) are true • xA(x) is true iff for all constants c, A(c) is true • A(c) --- the result of substituting all free occurrences of x by c in A(x) 10+10=1010 v(v+v=vv) v (v1=v) v((v+v=v)) v(v=0(v+v=v)) vv’(v+v’=v’+v) vv’(v(v’+1)=(v’v)+v)

  5. 6.2.d Giorgi JaparidzeTheory of Computability Let Th(N,+, ) = {A | A is a true arithmetical sentence} The undecidability of truth for arithmetic sentences Theorem 6.13: Th(N,+, ) is undecidable. Corollary:Th(N,+, ) is not Turing recognizable, either. Proof: Suppose a TM M recognizes Th(N,+, ). Construct a TM D: D = “On input A, and arithmetic sentence, 1. Run M on both A and A in parallel. 2. If M accepts A, accept; if M accepts A, reject” Obviously D decides Th(N,+, ) , which is in contradiction with Theorem 6.11. Let Th(N,+) = {A | A is a true arithmetical sentence not containing } Theorem 6.12: Th(N,+) is decidable.

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