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Proof step 2 from proof in-09 (previous step)

x + a ' = y + b ' + z, b ∈ z ⊦ dups(x + a ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x + a ' = y + b ' + z, b ∈ z, a ∈ y ∨ a ∈ z, b ∈ x ∨ b = a ⊦ dups(x), a ∈ x

Rule argument

Additional information

Used simplifier rules: f: ⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x);% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%in-13: ⊦ x = y + a ' + z → a ∈ x; (forward) from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list