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Proof step 3 from proof in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -l b ⊦ a = b, a = a0, a ∈ (a0 ' + x0) -l b

Rule argument

Additional information

Used simplifier rules: del-n: ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a);%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ a + x = a ' + x; from specification list