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Proof step 3 from proof first-app (previous step)

x + y = a0 + x0 ∧ x + y ≠ [], a ∈ x ⊦ a0 = x.first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
x + y = a0 ' + x0, a ∈ x, a0 ∈ x ∨ a0 ∈ y ⊦ x + y = [], a0 = x.first

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%in-14: ⊦ a ' + x = y → a ∈ y;%a: ⊦ (x + y) + z = x + y + z;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);