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Proof step 3 from proof dups (
previous step
)
dups(set2list(s)), ∀ s0. # s0 < # s → dups(set2list(s0)) → false ⊦ false
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 4
:
dups(set2list(s)), ∀ s0. # s0 < # s → dups(set2list(s0)) → false ⊦
Rule argument
Additional information