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Proof step 19 from proof get-nodups (previous step)

x0[m1] = x0[m0], m1 < # x0, m0 < # x0, ∀ m, n0. x0[m] = x0[n0] ∧ n0 < # x0 ∧ m < # x0 → m = n0 ⊦ x0 = [], m1 = m0, dups(x0), a ∈ x0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m1, m0

Additional information

Used terms: m1, m0