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Proof step 5 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 ' + x0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a ∈ y, a0 ∈ y, ∀ a, y. a ∈ y → (a ∈ filter(x0, y) ↔ a ∈ x0) ⊦ a ∈ filter(a0 ' + x0, y) ↔ a ∈ a0 ' + x0
Proof step 12:
a ∈ y, ∀ a, y. a ∈ y → (a ∈ filter(x0, y) ↔ a ∈ x0) ⊦ a ∈ filter(a0 ' + x0, y) ↔ a ∈ a0 ' + x0, a0 ∈ y

Rule argument

fma: a0 ∈ y

Additional information

Used formulas: a0 ∈ y