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Proof step 1 from proof two (conclusion)

⊦ a ' + b ' = x + a0 ' + c ' + y ↔ x = [] ∧ y = [] ∧ a = a0 ∧ b = c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ' + b ' = x + a0 ' + c ' + y, a ∈ x ∨ a = a0 ∨ a = c ∨ a ∈ y ⊦ x = [] ∧ y = [] ∧ a = a0 ∧ b = c

Rule argument

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%app: ⊦ x + [] = x;%a: ⊦ (x + y) + z = x + y + z;%app-08: ⊦ a ' + x = a ' + y ↔ x = y;%in-14: ⊦ a ' + x = y → a ∈ y;%in-15: ⊦ x + a ' = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%in-16: ⊦ x + a ' = y + b ' + z ∧ a ≠ b → a ∈ z;