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Proof step 3 from proof in-04 (previous step)

a ∈ x ⊦ a = b, a ∈ x -1l b

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
a ∈ a0 + x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -1l b ⊦ a = b, a ∈ (a0 + x0) -1l b

Rule argument

term: x

Additional information

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