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Proof step 3 from proof del1-empty (previous step)

x -1l a = [] ⊦ x = [], x = a '

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;