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. Proof, Standard properties of any non-substructural type system. Structural congruence alters the basic arrangement of processes without affecting typing

. Proof, By induction on the derivation of P ?? Q and by cases on the last rule applied

[. Then and P. , | vw and = ?. From [T-THREAD] we deduce ? E [fork v w] : unit. From Lemma 3 and TypeOf(fork) we deduce that ? fork v w : unit and ? v : t ? -unit and ? w : t. From Lemma 4 we deduce ? E [()] : unit. From [T-APP] we deduce ? vw : unit. We conclude with two applications of

[. Then and P. , where a is fresh and = ?. From [T-THREAD] we deduce ? E [create()] : unit. From Lemma 3 and TypeOf(create) we deduce ? create() : t * t ? . Since a is fresh we have a + , a -, a * ? dom(? ). From Lemma 4 we deduce ?

[. Then and P. , -THREAD] we deduce ? E [close a p ] : unit and ? E [close a p ] : unit. From Lemma 3 and TypeOf(close) we deduce ? (a p ) = ? (a p ) = ??, ?? [unit] for some ? such that ? ?? ? . From Lemma 4 we deduce ? E [()] : unit and ? E [()] : unit. We conclude with two applications of [T-THREAD] and one application of [T-PAR], p.??
URL : https://hal.archives-ouvertes.fr/hal-00505256

T. Then and P. , From the hypothesis ? P and rules [T-PAR] and [T-THREAD] we deduce that ? E [left a p ] : unit and ? E [branch a p ] : unit. From Lemma 3 and TypeOf(left) and the hypothesis that ? is balanced we deduce that there exists ? such that ? = ? , a p : ?? [t + s], a p : ?? [t + s] and ? left a p : t ? . From Lemma 3 and TypeOf(branch) we deduce that ? branch a p : t + s. Let ? def = ? , a p : t ? , a p : t and observe that ? ?? ? . From Lemma 4 we derive ? E [a p ] : unit. From one application of, We conclude with two applications of

?. If and =. Ca, By Lemma 5, we deduce ? , a + : t, a -: t ? , a * : ?A.A Q and we conclude by taking ? = ? and one application of

?. If and =. Map, then by induction hypothesis we deduce that there exists s such that ? , a + : t, a -: t ? , a * : ?A.A map ?? ? , a + : s, a -: s ? , a * : ?A.A and ? , a + : s, a -: s ? , a * : ?A.A Q