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Proof step 4 from proof eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ' + b ' + y + c ' + z = x0 + c ' + a ' + z0 ⊦ a = c, a = b, dups(z), c ∈ z, a ∈ z, a ∈ y

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ ¬ a ∈ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ [] + x = x; (s)