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Läßt sich (A) => A (für beliebiges A ∈ L) ableiten? Ja:
P1 | = | H1(A, (A) => A) |
= | (A) => ((A) => A) => A | |
P2 | = | H2(A, (A) => A, A) |
= | ((A) => ((A) => A) => A) => ((A) => (A) => A) => (A) => A | |
P3 | = | Modus Ponens(P1, P2) |
= | ((A) => (A) => A) => (A) => A | |
P4 | = | H1(A, A) |
= | (A) => (A) => A | |
P5 | = | Modus Ponens(P4, P3) |
= | (A) => A |
Copyright © 2004, 2005 Andreas Borchert, in HTML konvertiert am 18.02.2005 |