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Proof step 1 from proof inapp-05 (conclusion)

⊦ a ≠ b ∧ ¬ b ∈ x ∧ ¬ b ∈ x0 → (a ' + x + b ' + y = x0 + b ' + y0 ↔ a ' + x = x0 ∧ y = y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ' + x + b ' + y = x0 + b ' + y0, a ∈ x0 ⊦ b ∈ x, b ∈ x0, a ' + x = x0 ∧ y = y0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%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);%inapp-02: ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0);