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Proof step 2 from proof set2set (
previous step
)
true, ∀ s0. # s0 < # s → true → list2set(set2list(s0)) = s0 ⊦ list2set(set2list(s)) = s
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
∀ s0. # s0 < # s → true → list2set(set2list(s0)) = s0 ⊦ list2set(set2list(s)) = s
Rule argument
Additional information