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Logics and Type Systems
, 1993
"... from the last declaration in \Delta (which is p:'). (oeE) In fact the ([\Theta]) is not exactly the ([\Theta]) that is found by induction. Possibly some of the free variables in ([\Theta]) are renamed. What happens is the following: 1. Consider the proofcontext \Delta 1 ] \Delta 2 and especially ..."
Abstract

Cited by 84 (5 self)
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from the last declaration in \Delta (which is p:'). (oeE) In fact the ([\Theta]) is not exactly the ([\Theta]) that is found by induction. Possibly some of the free variables in ([\Theta]) are renamed. What happens is the following: 1. Consider the proofcontext \Delta 1 ] \Delta 2 and especially the renaming of the declared variables in \Delta 2 that has been caused by the operation ]. 2. Rename the free proofvariables in ([\Theta]) accordingly, obtaining say, ([\Theta 0 ]). 3. Apply ([\Sigma]) to ([\Theta 0 ]). (There will in practice be no confusion if we just write ([\Theta]) instead.) Of course the intended meaning is that the judgement below the double lines is derivable if the judgement above the lines is. This will be proved later in Theorem 3.2.8. It should be clear at this point however that there is a onetoone correspondence between the occurrences of ' as a (nondischarged) premise in the deduction and declarations p:' in \Delta. Notation. If for \Sigma a deducti...
The maximality of the typed lambda calculus and of cartesian closed categories
 Publ. Inst. Math. (N.S
"... From the analogue of Böhm’s Theorem proved for the typed lambda calculus, without product types and with them, it is inferred that every cartesian closed category that satisfies an equality between arrows not satisfied in free cartesian closed categories must be a preorder. A new proof is given here ..."
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Cited by 17 (2 self)
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From the analogue of Böhm’s Theorem proved for the typed lambda calculus, without product types and with them, it is inferred that every cartesian closed category that satisfies an equality between arrows not satisfied in free cartesian closed categories must be a preorder. A new proof is given here of these results, which were obtained previously by Richard Statman and Alex K. Simpson.