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Proof step 12 from proof rem-inter-empty3 (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 13:
a ∈ s1, a ∈ s2, ∀ a0. ¬ a0 ∈ s \ s0 ∨ ¬ a0 ∈ s1, ¬ a ∈ s2 \ s ∨ ¬ a ∈ s1 ⊦ # s1 = 0, # s2 = 0, #((s2 \ s0) ∩ s1) = 0, s1 = ∅, s2 = ∅, a ∈ s0

Rule argument

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

Additional information

Used terms: a