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Proof step 11 from proof rem-inter-empty3 (previous step)

a ∈ s2, a ∈ s, ∀ a0. ¬ a0 ∈ s2 \ s0 ∨ ¬ a0 ∈ s, ∀ a0. ¬ a0 ∈ s0 \ s1 ∨ ¬ a0 ∈ s ⊦ s2 = ∅, s = ∅, #((s2 \ s1) ∩ s) = 0, # s2 = 0, # s = 0, a ∈ s1

User applied rule all left

And obtained the following new goals

Proof step 12:
a ∈ s2, a ∈ s, ¬ a ∈ s2 \ s0 ∨ ¬ a ∈ s, ∀ a0. ¬ a0 ∈ s0 \ s1 ∨ ¬ a0 ∈ s ⊦ s2 = ∅, s = ∅, #((s2 \ s1) ∩ s) = 0, # s2 = 0, # s = 0, a ∈ s1

Rule argument

pos: left 3 terms: a abort: F discard: T computed: T

Additional information

Used terms: a