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Proof step 2 from proof app-06 (previous step)

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ x, a, b. x + a ' = y0 + b ' ∧ (a ∈ y0 ∨ a = b) ∧ (b ∈ x ∨ b = a) → x = y0 ∧ a = b, x + a ' = (a0 + y0) + b ', a ∈ a0 + y0 ∨ a = b, b ∈ x ∨ b = a ⊦ x = a0 + y0 ∧ a = b

Rule argument

term: y

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c-05: ⊦ a ' = x + b ' ↔ a = b ∧ x = [];%in-empty: ⊦ ¬ a ∈ [];%list freely generated by [], +;;