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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, b ∈ x ∨ b ∈ a0 ' ∨ b ∈ c ' ∨ b ∈ 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-07: ⊦ a ' + x = b ' + x ↔ a = b;%in-14: ⊦ a ' + x = y → a ∈ y;%in-15: ⊦ x + a ' = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;