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Proof step 1 from proof neq-07 (conclusion)

⊦ a ≠ a0 ∧ a1 ≠ a0 ∧ a ≠ a1 ∧ ¬ a ∈ y1 ∧ ¬ a1 ∈ y0 → y1 + a0 ' + y0 + a ' ≠ y2 + a ' + a1 ' + z0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
y1 + a0 ' + y0 + a ' = y2 + a ' + a1 ' + z0 ⊦ a = a0, a1 = a0, a = a1, a ∈ y1, a1 ∈ y0

Rule argument

Additional information

Used simplifier rules: in-16: ⊦ x + a ' = y + b ' + z ∧ a ≠ b → a ∈ z;%a: ⊦ (x + y) + z = x + y + z;%in-15: ⊦ x + a ' = y → a ∈ y;%in-16: ⊦ x + a ' = y + b ' + z ∧ a ≠ b → a ∈ z;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);