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Proof step 5 from proof in-04 (previous step)

a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -1l b ⊦ b = a0, a = b, a = a0, a ∈ (a0 ' + x0) -1l b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -1l b ⊦ b = a0, a = b, a = a0, a ∈ x0 -1l b

Rule argument

Additional information

Used simplifier rules: del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a);%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list