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Proof step 4 from proof in-01-01 (previous step)

a ∈ y, ∀ a, y. a ∈ y → (a ∈ filter(x0, y) ↔ a ∈ x0) ⊦ a ∈ filter(a0 ' + x0, y) ↔ a = a0 ∨ a ∈ x0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a0 ∈ y, a ∈ y, ∀ a, y. a ∈ y → (a ∈ filter(x0, y) ↔ a ∈ x0) ⊦ a ∈ filter(a0 ' + x0, y) ↔ a = a0 ∨ a ∈ x0
Proof step 11:
a ∈ y, ∀ a, y. a ∈ y → (a ∈ filter(x0, y) ↔ a ∈ x0) ⊦ a0 ∈ y, a ∈ filter(a0 ' + x0, y) ↔ a = a0 ∨ a ∈ x0

Rule argument

fma: a0 ∈ y

Additional information

Used formulas: a0 ∈ y