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Proof step 1 from proof in-02 (conclusion)

⊦ a ≠ b → (a ∈ b ' ∪ y ↔ a ∈ y)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: union: ⊦ x ∪ y = rmdup(x + y);%assoc: ⊦ (x ∪ y) ∪ z = x ∪ y ∪ z;%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%rm-in-02: ⊦ a ∈ rmdup(x) ↔ a ∈ x; from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%rm-in: ⊦ a ∈ rmdup(x + y) ↔ a ∈ x ∨ a ∈ y; from specification list-dup