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Proof step 3 from proof eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%in-empty: ⊦ ¬ a ∈ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%Nil: ⊦ [] + x = x; from specification list