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Proof step 2 from proof inapp-08 (previous step)

a ' + x = y + b ', a ∈ y ∨ a = b, b = a ∨ b ∈ x ⊦ y = [], a ∈ y ∧ b ∈ x

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: y

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%e-01: ⊦ a ' + x ≠ [];%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%in-15: ⊦ x + a ' = y → a ∈ y;%list freely generated by [], +;;