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Proof step 5 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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a0 ∈ y -1l a, ∀ 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
Proof step 8:
∀ 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

Rule argument

fma: a0 ∈ y -1l a

Additional information

Used formulas: a0 ∈ y -1l a