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Proof step 8 from proof del1 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ y, a. filter(x0, y -1l a) = filter(x0, y) ∨ a ∈ x0 ⊦ a = a0, filter(x0, y -1l a) = filter(x0, y), a ∈ x0, a0 ∈ y

Rule argument

Additional information

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