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Proof step 2 from proof inl (previous step)

a ∈ x, # x ≤ pos(a, x) ⊦

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a. ¬(a ∈ x0 ∧ # x0 ≤ pos(a, x0)), a ∈ a0 + x0, #(a0 + x0) ≤ pos(a, a0 + x0) ⊦

Rule argument

term: x

Additional information

Used simplifier rules: list freely generated by [], +;;%in-empty: ⊦ ¬ a ∈ []; from specification list