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Proof step 1 from proof inapp-07 (conclusion)

⊦ a ' + x = y + b ' ∧ x ≠ [] → a ∈ y ∧ b ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
b ' + x = y + b ', b ∈ x ⊦ x = [], b ∈ y

Rule argument

Additional information

Used simplifier rules: in-14: ⊦ a ' + x = y → a ∈ y;%a: ⊦ (x + y) + z = x + y + z;%in-15: ⊦ x + a ' = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-17: ⊦ a ∈ b ' ↔ a = b;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%ni: ⊦ x + a ' = y + z ∧ ¬ a ∈ z → z = [];%in-12: ⊦ a ∈ y → a ∈ x + y;