On the Non-truth-functionality of Can the Multiplicative Fragment of Linear Logic be shown to be non-truth-functional?

Forgive me for undoing the last edit. The previous changes made my "par" symbols (which should read like upside down ampersands) illegible.

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edited Oct 8, 2010 at 17:54

$[ϕ(A) = ϕ(B)] \implies [ϕ(\lnot A) = ϕ(\lnot B)]$ ϕ(A) = ϕ(B) ==> ϕ(~A) = ϕ(~B)
$[ϕ(A1) = ϕ(A2) \mbox{ and } ϕ(B1) = ϕ(B2)] \implies [ϕ(A1⊗B1) = ϕ(A2⊗B2)]$ ϕ(A1) = ϕ(A2) AND ϕ(B1) = ϕ(B2) ==> ϕ(A1 ⊗ B1) = ϕ(A2 ⊗ B2)
$[ϕ(A1) = ϕ(A2) \mbox{ and } ϕ(B1) = ϕ(B2)] \implies [ϕ(A1⅋B1) = ϕ(A2⅋B2)]$ ϕ(A1) = ϕ(A2) AND ϕ(B1) = ϕ(B2) ==> ϕ(A1 ⅋ B1) = ϕ(A2 ⅋ B2)

ϕ(A1) = ϕ(A2) = e ϕ(B1) = ϕ(B2) = e

improved formatting somewhat

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edited Oct 8, 2010 at 8:23

(1) [ϕ(A) = ϕ(B)] ==> [ϕ(~A) = ϕ(~B)]

(2) [ϕ(A1) = ϕ(A2) and ϕ(B1) = ϕ(B2)] ==> [ϕ(A1⊗B1) = ϕ(A2⊗B2)] (3) [ϕ(A1) = ϕ(A2) and ϕ(B1) = ϕ(B2)] ==> [ϕ(A1⅋B1) = ϕ(A2⅋B2)]

ϕ(A1) = ϕ(A2) = e ϕ(B1) = ϕ(B2) = e

deleted 67 characters in body

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edited Oct 7, 2010 at 22:26

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asked Oct 7, 2010 at 22:20

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