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Proof step 2 from proof inapp-07 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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;%ni: ⊦ x + a ' = y + z ∧ ¬ a ∈ z → z = [];