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Proof step 7 from proof two (previous step)

a ' + b ' = x + a0 ' + c ' + a1 ' + y0, a ∈ x, ∀ a, b, x, a0, c. a ' + b ' = x + a0 ' + c ' + y0 ∧ (a ∈ x ∨ a = a0 ∨ a = c ∨ a ∈ y0) → x = [] ∧ y0 = [] ∧ a = a0 ∧ b = c ⊦ b ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ' + b ' = x + a0 ' + c ' + a1 ' + y0, a ∈ x, ∀ a, b, x, a0, c. a ' + b ' = x + a0 ' + c ' + y0 ∧ (a ∈ x ∨ a = a0 ∨ a = c ∨ a ∈ y0) → x = [] ∧ y0 = [] ∧ a = a0 ∧ b = c ⊦ b ∈ x

Proof step 7 from proof two (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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