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Proof step 5 from proof del1 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ y, ∀ y, a. filter(x0, y -1l a) = filter(x0, y) ∨ a ∈ x0 ⊦ a = a0, filter(x0, y -1l a) = filter(x0, y), a ∈ x0

Rule argument

Additional information

Used simplifier rules: filt-y: ⊦ a ∈ y → filter(a ' + x, y) = a ' + filter(x, y);%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%del1-in: ⊦ a ≠ b → (a ∈ x -1l b ↔ a ∈ x); from specification list-del%a: ⊦ (x + y) + z = x + y + z; from specification list