CS 641 Lecture -*- Outline -*-

* Derivations
   Proof techniques, not specific to boolean lattices

** Assumptions (6.5)

   Assumptions and implications are related

          A sequent Phi |- t says that t follows from Phi.
          Assumptions are just a way to work with implications.

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       RULES FOR ASSUMPTIONS (6.5)

  Phi, t |- t'
 ________________      (discharge)
  Phi |- t ==> t'

  Phi |- t ==> t'
 ________________      (undischarge)
  Phi, t |- t'

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        Q: How would you prove the assume axiom,  Phi, t |- t ?
           use reflexivity and undischarge (in a derivation)

        Q: How does this let you use lemmas?

           Phi |- t   Phi', t |- t'
           ________________________        (use of lemmas, or cut rule)
             Phi \union Phi' | t'

           proof uses modus ponens and discharging

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         CASE ANALYSIS

 Phi, t |- t'  Phi', not.t |- t'
 __________________________(case analysis)
   Phi \union Phi' | t'
 

Proof. Assume Phi, t |- t'
and Phi', not.t |- t'.

  Phi \union Phi'
  |- T
  <==> { not exhaustion }
     t \/ not.t
  ==> { \/ elimination }
        * t
        ==> { undischarge, assumption }
          t'

        * not.t
        ==> { undischarge, assumption }
          t'
  . t'
------------------------------------------

        Note the mix of <==> and ==>
             mix yields the composition of the relations involved.
             Ok if all arrows go the same way.

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         MONOTONICITY AND ALPHA

 Phi |- t
 ___________             (add assumption)
 Phi, t' |- t

 Phi, s |- t
 _______________(change of free variables)
 Phi, s[x := x'] |- t[x :- x']
 * x is not free in Phi,
   and x' is free for x in s and t

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        Q: Why do these make sense?

** Derivations with local assumptions (6.6)

*** focusing rules for connectives
    The reason for using the focusing rules instead of just
    substituting equals for equals is to get the extra assumption.

------------------------------------------
            FOCUSING RULES
          UNDER EQUIVALENCE

Focusing on a conjunct:

    Phi, t |- t1 <==> t2
    ___________________________
    Phi |- t /\ t1 <==> t /\ t2

Example:

    |-   x = 3 /\ "y = x"
    <==> { focus on conjunct }
           * [ x = 3 ]
                y = x
           <==> { by assumption, x = 3 }
                y = 3
    .    x = 3 /\ y = 3
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        Q: why is this justified?
           because both conjuncts must be true

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   FOCUSING RULES FOR OTHER CONNECTIVES
           UNDER EQUIVALENCE

Focusing on a disjunct:

  Phi, not.t |- t1 <==> t2
  ___________________________
  Phi |- t \/ t1 <==> t \/ t2


Focusing on consequent of implication:

  Phi, t |- t1 <==> t2
  _____________________________
  Phi |- t ==> t1 <==> t ==> t2


Focusing on antecedent of an implication:

  Phi, not.t |- t1 <==> t2
  _____________________________
  Phi |- t1 ==> t <==> t2 ==> t

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        Q: How would you explain these?

        Q: Can you prove that p ==> ((p ==> q) ==> q) ?

   |-   p ==> ("(p ==> q) ==> q")
   <==> { focus on conclusion of ==> }
          * [ p ]
               (p ==> q) ==> q
          <==> { by assumption p <==> T }
               "(T ==> q)" ==> q
          <==> { T ==> rule }
               q ==> q
          <==> { ==> reflexive }
               T
    .   p ==> T
    <==> { T greatest }
        T

*** focusing and monotonicity
------------------------------------------
        FOCUSING AND MONOTONICITY
           (UNDER RELATIONS)

For monotonic functions we have:

   Phi |- t <= t'
   __________________
   Phi |- f.t <= f.t'
   * if f is monotonic

Can manipulate t separately.

Works inside quantifiers,
  since they are monotonic
  with respect to implication.

Example, prove:

   |- (\exists v * v = t /\ t')
        ==> t'[v := t]
   * if v is not free in t
------------------------------------------
        Note, no extra hypothesis here.

        Q: What happens if f is antimonotonic?

   Proof: Assume v is not free in t.

   |-   (\exists v * "v = t /\ t'")
   <==> { substitute using local assumption v = t }  // note, no subderivation!
        (\exists v * "v = t /\ t'[v := t]")
   ==>  { /\ elimination in monotonic context }
        (\exists v * "t'[v := t]")in the
   <==> { drop vacuous quantifier, exercise 6.8; v is not free in t }
        t'[v := t]

  Note the abbreviation where the comment is.code

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         FOCUSING UNDER IMPLICATION

Focusing on a conjunct (exercise 6.7a)

        Phi, t |- t1 ==> t2
      _____________________________
       Phi |- t /\ t1 ==> t /\ t2


Focusing on a disjunct (exercise 6.7b)




Focusing on the consequent
of an implication (exercise 6.7b)




Focusing on the antecedent
of an implication (exercise 6.7b)




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        the formulation is the dual of the rule for conjunction

        Q: What's a monotonic context?
        Q: What parts of an implication are monotonic contexts?

*** focusing and local definitions
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  USING LOCAL DEFINITIONS AS ASSUMPTIONS

   Phi, v = t |- t1 ~ t2
   __________________________
   Phi |- (let v = t in t1)
          ~ (let v = t in t2)

   where ~ is an arbitrary relation
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        See procedure example on p. 123

        Get to use definition as another assumption in subproof.