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Proof step 8 from proof inapp (previous step)

z = a0 + y1 ∧ z ≠ [], x0 + a ' + y = (z + a ').rest + y0, (z + a ').first ∈ z, ∀ a, y, z, y0. x0 + a ' + y = z + a ' + y0 → a ∈ y ∨ a ∈ x0 ∨ y = y0 ∧ x0 = z ⊦ a = a0, a ∈ x0, a ∈ y, y = y0 ∧ a0 ' + x0 = z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
x0 + a ' + y = (y1 + a ') + y0, ∀ a, y, z, y0. x0 + a ' + y = z + a ' + y0 → a ∈ y ∨ a ∈ x0 ∨ y = y0 ∧ x0 = z ⊦ a = a0, a ∈ x0, a ∈ y, y = y0 ∧ x0 = y1

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-07: ⊦ (a ' + x).rest = x;%e-01: ⊦ a ' + x ≠ [];%c-08: ⊦ (a ' + x).first = a;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%app-08: ⊦ a ' + x = a ' + y ↔ x = y;