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Proof step 14 from proof diff-unchanged (previous step)

∀ y. x0 \ y = x0 ↔ (∀ a. ¬ a ∈ x0 ∨ ¬ a ∈ y), ∀ a0. ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y ⊦ x0 \ y = x0, a ∈ y

User applied rule all left

And obtained the following new goals

Proof step 15:
x0 \ y = x0 ↔ (∀ a. ¬ a ∈ x0 ∨ ¬ a ∈ y), ∀ a0. ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y ⊦ x0 \ y = x0, a ∈ y

Rule argument

pos: left 1 terms: y abort: F discard: T computed: T

Additional information

Used terms: y