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Proof step 6 from proof neq-07 (previous step)

∀ a0, y0, y2, a2, a1, y. y3 + a0 ' + y0 = y2 + a2 ' + a1 ' + y → a2 = a1 ∨ a1 = a0 ∨ a2 = a0 ∨ a1 ∈ y0 ∨ a2 ∈ y3, (a + y3) + a0 ' + y0 = y2 + a2 ' + a1 ' + y ⊦ a2 = a1, a1 = a0, a2 = a0, a1 ∈ y0, a2 ∈ a + y3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(a ' + y3) + a0 ' + y0 = y2 + a2 ' + a1 ' + y, a ∈ y2, ∀ a0, y0, y2, a2, a1, y. y3 + a0 ' + y0 = y2 + a2 ' + a1 ' + y → a2 = a1 ∨ a1 = a0 ∨ a2 = a0 ∨ a1 ∈ y0 ∨ a2 ∈ y3 ⊦ a2 = a, a2 = a0, a1 = a0, a2 = a1, a2 ∈ y3, a1 ∈ y0

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x;%in-14: ⊦ a ' + x = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x);%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%inapp-02: ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0);