Connectivity Corollary. GRAPH CONNECTIVITY is not FO definable
Connectivity Corollary. GRAPH CONNECTIVITY is not FO definable • If A is a linear order of size n, let G( A ) be the graph with edges { i, i+2 mod n } for all a ∈ A
Connectivity Corollary. GRAPH CONNECTIVITY is not FO definable • If A is a linear order of size n, let G( A ) be the graph with edges { i, i+2 mod n } for all a ∈ A
Connectivity ∉ FO Obs 1: G( A ) is connected if and only if n is odd Corollary. GRAPH CONNECTIVITY is not FO definable • If A is a linear order of size n, let G( A ) be the graph with edges { i, i+2 mod n } for all a ∈ A
Connectivity ∉ FO Obs 2: G( A ) is first-order definable from A Corollary. GRAPH CONNECTIVITY is not FO definable • If A is a linear order of size n, let G( A ) be the graph with edges { i, i+2 mod n } for all a ∈ A
Connectivity Corollary. GRAPH CONNECTIVITY is not FO definable • If A is a linear order of size n, let G( A ) be the graph with edges { i, i+2 mod n } for all a ∈ A • If 𝜒 were a first-order formula defining GRAPH CONNECTIVITY, then by replacing each sub-formula E(x,y) with a formula “x and y have cyclic distance 2 in the linear order A ”, we could define EVENNESS of A (which we showed is impossible by the EF game).
Connectivity Corollary. GRAPH CONNECTIVITY is not FO definable • This result can be proved directly by playing the EF game e.g. on graphs C n and C n + C n • The reduction to EVENNESS of linear orders illustrates the technique of a first-order interpretations .
Set-powersets • SetPow n is the structure ([n] ∪ 2 [n] , Atoms, Sets, In) where Atoms = [n] = {1,...,n}, Sets = powerset of Atoms, In = {(i,X) ∈ Atoms × Sets | i ∈ X}.
Set-powersets • SetPow n is the structure ([n] ∪ 2 [n] , Atoms, Sets, In) where Atoms = [n] = {1,...,n}, Sets = powerset of Atoms, In = {(i,X) ∈ Atoms × Sets | i ∈ X}. • A set-powerset is any structure A with relations {Atoms, Sets, In} which is isomorphic to SetPow n for some n > 0. It is said to be EVEN/ODD according to the parity of n.
Set-powersets Obs. The class of set-powersets is FO definable. • We cannot say (in first-order logic): ∀ X ⊆ Atoms ∃ S ∈ Sets ∀ x ∈ Atoms, x ∈ X ⇔ In(x,S) • Instead, we say: " ⦰ ∈ Sets" ∧ ∀ S ∈ Sets ∀ x ∈ Atoms "S ∪ {x} ∈ Sets" • This formula exploits finiteness in an essential way.
Set-powersets Theorem The class of EVEN set-powersets is not FO definable.
Set-powersets Theorem The class of EVEN set-powersets is not FO definable. Proof For every k, we show that Duplicator has a winning strategy in the k-round Ehrenfeucht-Fraisse game on A = SetPow 2^k and B = SetPow 2^k+1
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms Sets Sets
Winning strategy (by picture) A B Atoms Atoms At this point, Spoiler wins. Sets Sets
Winning strategy (by picture) Duplicator has a winning strategy A B for k rounds provided both structures have ≥ 2 k atoms. Atoms Atoms At this point, Spoiler wins. Sets Sets
Winning strategy (by picture) Duplicator has a winning strategy A B for k rounds provided both structures have ≥ 2 k atoms. Atoms Atoms At this point, Spoiler wins. Duplicator’s winning strategy: in round j, preserve the cardinality up to 2 k−j of every Boolean combination of the chosen sets Sets Sets and atoms
0-1 Laws
G(n,p) • The Erdos-Renyi random graph G(n,p) – vertex set {1,...,n} – indep. edge probability p
G(n,p) • The Erdos-Renyi random graph G(n,p) – vertex set {1,...,n} – indep. edge probability p G(n,½) is known as the uniform random graph
0-1 Law for FO Theorem [Fagin 1976, Glebskii et al 1969] For every FO sentence ϕ, lim n→∞ Pr[ G(n,½) ⊨ ϕ ] ∈ {0,1}
0-1 Law for FO Theorem [Fagin 1976, Glebskii et al 1969] For every FO sentence ϕ, lim n→∞ Pr[ G(n,½) ⊨ ϕ ] ∈ {0,1} That is, ϕ is either almost surely true or almost surely false in G(n,½)
0-1 Law for FO Theorem [Fagin 1976, Glebskii et al 1969] For every FO sentence ϕ and p ∈ (0,1), lim n→∞ Pr[ G(n,p) ⊨ ϕ ] ∈ {0,1}
0-1 Law for FO Theorem [Spencer and Shelah 1988] For every FO sentence ϕ and irrational c ∈ (0,1), lim n→∞ Pr[ G(n,n −c ) ⊨ ϕ ] ∈ {0,1}
0-1 Law for FO Theorem [Spencer and Shelah 1988] For every FO sentence ϕ and irrational c ∈ (0,1), lim n→∞ Pr[ G(n,n −c ) ⊨ ϕ ] ∈ {0,1} No 0-1 law for rational c ∈ (0,1)
Proof of The Zero-One Law
0-1 Law for FO Theorem [Fagin 1976, Glebskii et al 1969] For every FO sentence ϕ, lim n→∞ Pr[ G(n,½) ⊨ ϕ ] ∈ {0,1}
0-1 Law for FO • The 2-extension property EXT 2 says: Every vertex has a neighbor and a non-neighbor
0-1 Law for FO • The 3-extension property EXT 3 says: Every two vertices and have – a common neighbor – a common non-neighbor – a neighbor of each one, but not the other
0-1 Law for FO • The 4-extension property EXT 4 says: For all distinct vertices , there exist 8 vertices witnessing each possible set of adjacencies
0-1 Law for FO • The k-extension property EXT k says: for all k−1 distinct vertices, there exist 2 k witnesses. ……
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma If G and H satisfy EXT k , then Duplicator has a winning strategy in the k-round EF game on G and H (simply by maintaining a partial isomorphism)
0-1 Law for FO Lemma For every k, the uniform random graph G ～ G(n,½) satisfies EXT k a.a.s.
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