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Proof step 2 from proof eq-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x + a ' + b ' + y + c ' + z = x0 + a ' + c ' + z0 ⊦ a = b, a = c, b = c, dups(x + z), a ∈ y, a ∈ x, a ∈ z, c ∈ x, c ∈ 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% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)