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C1: ∀xWeakPref(x,x)

C2: ∀xIndiff(x,x)

C3: ∀x∀y(Indiff(x,y)↔Indiff(y,x))

C4: ∀x∀y∀z((Indiff(x,y)∧Indiff(y,z))→Indiff(x,z))

C5: ∀x∀y(StrongPref(x,y)→WeakPref(x,y))

C6: ∀x∀y(StrongPref(x,y)→ ¬StrongPref(y,x))

C7: ∀x∀y∀z((StrongPref(x,y)∧StrongPref(y,z))→StrongPref(x,z))

C8: ∀x∀y(Indiff(x,y)→¬(StrongPref(y,x)∨StrongPref(x,y)))

C9: ∀x∀y∀z((Indiff(x,y)∧StrongPref(y,z))→StrongPref(x,z))

C10: ∀x∀y∀z((Indiff(x,y)∧StrongPref(z,x))→StrongPref(z,y))

Daniel J.
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1 Answers1

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You only need premises P7 and P11. In order to give you a sketch of a proof, I will leave out the universal quantifiers: you will need to put in the universal instantiation and universal generalisation steps.

From P7 you have Indiff(x,x) and hence by disjunction introduction StrongPref(x,x) ∨ Indiff(x,x)

From P11, by substitution x/y you have WeakPref(x,x) ↔ (StrongPref(x,x) ∨ Indiff(x,x))

Hence (StrongPref(x,x) ∨ Indiff(x,x)) → WeakPref(x,x)

and by MP you have WeakPref(x,x)

Bumble
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