How much arithmetic is required to formalize quantifier elimination in Presburger arithmetic?
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As we know, Presburger arithmetic can be proved decidable by demonstrating that it admits quantifier elimination, i.e. that there is an algorithm that reduces any sentence in the language to some equivalent quantifier-free sentence (which in turn can be decided). As a consequence, Presburger arithmetic is consistent, because it can easily be computed that 0 = 1 is not a theorem.
I am wondering how much first-order arithmetic is required to prove Presburger arithmetic admits quantifier elimination. We know Robinson's Q is not strong enough to prove Con(Presburger); how much more is required?
logic proof-theory reverse-math quantifier-elimination presburger-arithmetic
edited Dec 8 '18 at 16:24
Rodrigo de Azevedo
13k41958
asked Oct 27 '18 at 2:40
Morgan SinclaireMorgan Sinclaire
385
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add a comment |
$begingroup$
As we know, Presburger arithmetic can be proved decidable by demonstrating that it admits quantifier elimination, i.e. that there is an algorithm that reduces any sentence in the language to some equivalent quantifier-free sentence (which in turn can be decided). As a consequence, Presburger arithmetic is consistent, because it can easily be computed that 0 = 1 is not a theorem.
I am wondering how much first-order arithmetic is required to prove Presburger arithmetic admits quantifier elimination. We know Robinson's Q is not strong enough to prove Con(Presburger); how much more is required?
logic proof-theory reverse-math quantifier-elimination presburger-arithmetic
edited Dec 8 '18 at 16:24
Rodrigo de Azevedo
13k41958
asked Oct 27 '18 at 2:40
Morgan SinclaireMorgan Sinclaire
385
$endgroup$
add a comment |
$begingroup$
As we know, Presburger arithmetic can be proved decidable by demonstrating that it admits quantifier elimination, i.e. that there is an algorithm that reduces any sentence in the language to some equivalent quantifier-free sentence (which in turn can be decided). As a consequence, Presburger arithmetic is consistent, because it can easily be computed that 0 = 1 is not a theorem.
I am wondering how much first-order arithmetic is required to prove Presburger arithmetic admits quantifier elimination. We know Robinson's Q is not strong enough to prove Con(Presburger); how much more is required?
logic proof-theory reverse-math quantifier-elimination presburger-arithmetic
edited Dec 8 '18 at 16:24
Rodrigo de Azevedo
13k41958
asked Oct 27 '18 at 2:40
Morgan SinclaireMorgan Sinclaire
385
$endgroup$
As we know, Presburger arithmetic can be proved decidable by demonstrating that it admits quantifier elimination, i.e. that there is an algorithm that reduces any sentence in the language to some equivalent quantifier-free sentence (which in turn can be decided). As a consequence, Presburger arithmetic is consistent, because it can easily be computed that 0 = 1 is not a theorem.
I am wondering how much first-order arithmetic is required to prove Presburger arithmetic admits quantifier elimination. We know Robinson's Q is not strong enough to prove Con(Presburger); how much more is required?
logic proof-theory reverse-math quantifier-elimination presburger-arithmetic
logic proof-theory reverse-math quantifier-elimination presburger-arithmetic
edited Dec 8 '18 at 16:24
Rodrigo de Azevedo
13k41958
asked Oct 27 '18 at 2:40
Morgan SinclaireMorgan Sinclaire
385
edited Dec 8 '18 at 16:24
Rodrigo de Azevedo
13k41958
asked Oct 27 '18 at 2:40
Morgan SinclaireMorgan Sinclaire
385
edited Dec 8 '18 at 16:24
Rodrigo de Azevedo
13k41958
edited Dec 8 '18 at 16:24
Rodrigo de Azevedo
13k41958
edited Dec 8 '18 at 16:24
Rodrigo de Azevedo
13k41958
13k41958
asked Oct 27 '18 at 2:40
Morgan SinclaireMorgan Sinclaire
385
asked Oct 27 '18 at 2:40
Morgan SinclaireMorgan Sinclaire
385
asked Oct 27 '18 at 2:40
Morgan SinclaireMorgan Sinclaire
385
385
add a comment |
add a comment |
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