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Proof step 8 from proof subset-del (previous step)

a ∈ x → a ∈ y, ∀ a. a ∈ x → a ∈ y ⊦ ¬(∀ a0. a0 ∈ x → a0 ∈ y -l a) ↔ a ∈ x

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 9:
¬(∀ a0. a0 ∈ x → a0 ∈ y -l a), a ∈ x → a ∈ y, ∀ a. a ∈ x → a ∈ y ⊦ a ∈ x
Proof step 11:
a ∈ x, a ∈ x → a ∈ y, ∀ a. a ∈ x → a ∈ y ⊦ ¬(∀ a0. a0 ∈ x → a0 ∈ y -l a)

Rule argument

pos: right 1

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