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Proof step 1 from proof eq-01 (conclusion)

⊦ x + a ' + b ' + y + c ' + z = x0 + a ' + c ' + z0 ∧ ¬ a ∈ y ∧ a ≠ b ∧ b ≠ c → dups(x + a ' + c ' + z)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x + a ' + b ' + y + c ' + z = x0 + a ' + c ' + z0 ⊦ b = c, a = b, a = c, a ∈ y, a ∈ x, a ∈ z, c ∈ x, c ∈ z, dups(x + z)

Rule argument

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list