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Proof step 2 from proof inapp-07 (previous step)

b ' + x = y + b ', b ∈ x ⊦ x = [], b ∈ y

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: y

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%one-01: ⊦ y ≠ [] → (a ' = x + y ↔ x = [] ∧ a ' = y);%c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x;%list freely generated by [], +;;