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Proof step 9 from proof diffone-nodupssize (previous step)

a ∈ x0, ∀ m0, a1. # x0 = m0 + 1 ∧ a1 ∈ x0 → #(x0 \ a1 ') = m0 ∨ x0 = [] ⊦ x0 = [], #(x0 \ a ') + 1 = # x0, dups(x0), a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information