Decidability of logical theories Giorgi Japaridze Theory of Computability Decidability of logical theories Section 6.2

The language of arithmetic: formulas Giorgi Japaridze Theory of Computability Formulas --- strings produced by the following CFG: 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

The language of arithmetic: sentences 6.2.b Giorgi Japaridze Theory 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. 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))

Truth of arithmetic sentences Giorgi Japaridze Theory of Computability 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)

The undecidability of truth for arithmetic sentences Giorgi Japaridze Theory of Computability Let Th(N,+, ) = {A | A is a true arithmetical sentence} 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, an 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.