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Proof step 12 from proof del1-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
a0 ∈ y, ∀ y, a. filter(rmdup(x0), y -1l a) = filter(rmdup(x0), y) ∨ a ∈ x0 ⊦ a = a0, filter(rmdup(x0), y -1l a) = filter(rmdup(x0), y), a0 ∈ x0, a ∈ x0

Rule argument

Additional information

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