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Proof step 3 from proof valid-eraseqd-nodups1 (previous step)

valid-eraseq(q0 + q1), m ∈ q1, m ∈ q0 ⊦

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
valid-eraseq((eqent + q) + q1), m ∈ eqent + q, m ∈ q1, ∀ q1, m. ¬(valid-eraseq(q + q1) ∧ m ∈ q1 ∧ m ∈ q) ⊦

Rule argument

term: q0

Additional information

Used simplifier rules: queue freely generated by [], +;;% ⊦ ¬ n ∈ []; (s)