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Proof step 2 from proof first-02 (previous step)

x + a ' + y = a0 + y0 ∧ x + a ' + y ≠ [] ⊦ a0 = (x + a ').first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x + a ' + y = a0 ' + y0, a0 ∈ x ∨ a0 = a ∨ a0 ∈ y ⊦ a0 = (x + a ').first

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%e-03: ⊦ x + a ' + y ≠ [];%a: ⊦ (x + y) + z = x + y + z;%in-14: ⊦ a ' + x = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);