Download Gödel Completeness Incompleteness Theorems

Paradoxes?
Lecture 39: Gödel
Incompleteness Theorem
CSCI 81
Spring, 2012
• This statement is not true.
• This statement is not provable.
• Second is at heart of Gödel Incompleteness.
Kim Bruce
Gödel Completeness
Incompleteness Theorems
• Recall that T is consistent iff no false statement
• Different notion of completeness -- w.r.t. model
has a valid proof. (T ⊢φ ⇒ T ⊨φ).
• T is complete iff every true statement has a
valid proof. (T ⊨φ ⇒ T ⊢φ).
• Gödel Completeness:
In predicate logic, if T ⊨ φ, then T ⊢ φ.
• Gödel Incompleteness 1:
For every
“interesting” system there are true statements
that cannot be proved.
• Gödel Incompleteness 2:
For every
“interesting” system, the consistency of that
system cannot be proved within itself.
Peano Arithmetic
Completeness & Consistency
• Axioms:
• Every provable statement of PA is true of the
• PA0. ∀x ¬ 0 = s(x)
natural numbers.
• PA1. ∀x ∀y (s(x) = s(y) → x = y)
• What about completeness?
• PA2. ∀x (x + 0 = x)
• Is PA enough to prove all true statements in N?
• PA3. ∀x∀y(x+s(y)=s(x+y))
• Thm:
The set of statements provable from PA
is semidecidable.
• PA4. ∀x(x·0=0)
• PA5. ∀x∀y(x·s(y)=(x·y)+x)
• PA6. φ[0/x] ∧ ∀x (φ → φ[s(x)/x]) → ∀x φ
Schema
Incompleteness
• Let Th(N) = set of all sentences in language of
PA that are true in N
• Lemma:
Th(N) is not semi-decidable. Thus
{φ : PA ⊢ φ} ⊂ Th(N), but not equal.
• Proof:
Show ¬ HTM ≤M Th(N)
• Given <M,w>, construct γ s.t.
<M,w> ∈ ¬HTM iff γ ∈ Th(N)
• Already showed predicate logic not decidable.
Constructing γ
• Can encode TM computations as integers:
• Give characters an integer code and code sequence as
2c1 3c2 5c3 7c4 11c5 13c6 17c7 ...
• Can write formula ValidCompM,x(y) that says y represents
a valid computation history of M on input x.
• Define γ = ¬∃y ValidCompM,x(y) says M,x not halt!
• As desired, get <M,w> ∈ ¬HTM iff γ ∈ Th(N)
• Thus Th(N) not semidecidable, so {φ : PA ⊢ φ} ⊂ Th(N)
and hence PA incomplete -- can’t prove all φ true in N.
Gödel Incompleteness
• Gödel 1:
Let T be a decidable set of axioms
true of the natural numbers & that implies the
axioms of Peano Arithmetic. Then there is a
sentence γ which is true of N but is not
provable in T.
• Proof only depended on ability to encode computation.
• Set of statements provable from a decidable T is semidecidable, but Th(N) is not.
• T consistent ⇒ Provable(T) = {φ | T ⊢ φ} ⊆ Th(N).
Construct φ as Fixed-Point
• For any formula ψ, let ‘ψ’ be a numeric
encoding of ψ. Write φn for the formula with
numeric code n. Thus φ‘ψ’ = ψ
• Thm:
For any ψ(x) with one free variable x,
there exists a sentence τ s.t. ⊢PA τ ↔ ψ( ‘τ’)
Gödel’s Proof
• More interesting in that used fixed point to
construct sentence that asserts its own
unprovability. N ⊨ φ iff not ⊢PA φ.
• Suppose had such sentence.
Claim N ⊨ φ.
• Spose not: If N ⊭ φ then ⊢PA φ. But PA consistent, so
can’t prove false things. Thus N ⊨ φ.
• But then not ⊢PA φ, so φ is true but not
provable.
Encoding
• Construct Subst(x,y,z) asserting:
• z is the code of the formula obtained by substituting the
constant whose value is x for all free occurrences of the
variable x0 in the formula whose code is y.
• Subst(x,y,z) true iff φz = φy[x/x0]
• Thus Subst(47, ‘φ(x0)’, 292) true iff ‘φ(47)’ = 292.
Fixed Point Proof
Proving Incompleteness
• Can define formula Proof(x,y) true iff sequence
• Define
encoded by x is a valid proof in PA of φy.
• σ(x) = ∀y (Subst(x,x,y) → ψ(y))
• τ = σ( ‘σ(x0)’) gives the fixed point of ψ.
• τ = σ( ‘σ(x0)’) = ∀y (Subst(‘σ(x0)’, ‘σ(x0)’,y) → ψ(y))
↔ ∀y. ((y = ‘σ(‘σ(x0)’)’) → ψ(y))
= ∀y. (y = ‘τ’) → ψ(y)
↔ ψ(‘τ’)
• Entire argument can be formalized in PA so
⊢PA τ ↔ ψ( ‘τ’)
• i.e. Proof(‘π’, ‘φ’) iff π is a proof in PA of φ
• Define Provable(y) iff ∃x Proof(x,y)
• Thus ⊢PA φ iff N ⊨ Provable( ‘φ’ )
(*)
• In fact ⊢PA φ iff ⊢PA Provable( ‘φ’ ) because argument
can be encoded in PA.
• Use fixed point to get τ s.t
⊢PA τ ↔ ¬Provable(‘τ’)
Incompleteness
• ⊢PA τ ↔ ¬Provable(‘τ’)
• As above τ must be true and therefore not
provable:
• N⊨ τ ⇒ N⊨ ¬Provable(‘τ’) ⇒ N ⊭ Provable(‘τ’)
⇒ ⊬PA τ by * on previous page.
• Proved there is τ s.t. τ is true of N, but not
provable in PA.
Defining Truth?
• Can we define predicate True(x) s.t. for all φ,
⊨ φ iff ⊨ True( ‘φ’ ) ?
• By fixed point theorem, exists σ s.t.
• ⊨ σ iff ⊨ ¬True( ‘σ’ )
• Can reason about consistency and provability,
but not truth.
Consistency is Contradictory!
Proof of Gödel 2
• Let τ be fixed point:
• Define Consis = ¬ Provable(‘⊥’)
• Gödel’s second incompleteness theorem:
No
sufficiently powerful deductive system can
prove its own consistency, unless it is
inconsistent (and hence can prove anything).
• We’ll do proof with PA, but stronger works too.
⊢PA τ ↔ ¬Provable(‘τ’)
• If ⊢PA τ, then ⊢PA Provable(‘τ’), but by above also get
⊢PA ¬Provable(‘τ’), so inconsistent. (*)
• Formalizing in PA, get
⊢PA Provable(‘τ’) → ¬Consis
or equivalently,
⊢PA Consis → ¬Provable(‘τ’)
• Suppose ⊢PA Consis. Then ⊢PA ¬Provable(‘τ’) and hence,
⊢PA τ, and by (*), PA is inconsistent.
• Thus PA consistent ⇒ ¬⊢PA Consis
Alternate Proof of Gödel 1
using Recursion Theorem
• Define TM S as follows:
• Ignore input, and write description of self <S>
• Search (possibly forever) for a proof of ψ = ¬ AcceptsS,o
• Let T be a decidable set of axioms true of the
natural numbers & that implies the axioms of
Peano Arithmetic.
• Recall defined formula ∃y ValidCompM,x(y) true iff M
halts on x. Abbreviate it as AcceptsM,x
• Use recursion theorem and HaltsM,x to find fixed formula
that is unprovable.
from T.
• If find a proof then accept. If rejects then reject.
• Now,
• ψ true iff S not accept o
• If S finds proof of ψ then S accepts o, so S false!
• But T consistent, so S not find a proof of ψ, so S not
accept 0
• Hence ψ true but not provable.
If work even Harder
• Gödel 2nd Incompleteness:
No sufficiently
powerful deductive system can prove its own
consistency, unless it is inconsistent.
• Holds of number theory, set theory, etc.
• Destroyed Hilbert’s program.