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Proof step 4 from proof in-02-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a. x0 -1l a = x0 ∨ a ∈ x0 ⊦ x0 -1l a = x0, a = a0, a ∈ x0

Rule argument

Additional information

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