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Proof step 4 from proof in-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -1l b ⊦ a = b, a = a0, a ∈ (a0 ' + x0) -1l b

Rule argument

Additional information

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