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Proof step 13 from proof subseteq-rem-del (previous step)

a ∈ s0, a ∈ s1, ∀ a. a ∈ s0 → a ∈ s1, a0 = a ∨ a0 ∈ s1 ∧ ¬ a0 ∈ s0 ⊦ # s0 = 0, # s1 = 0, s0 = ∅, s1 = ∅, a0 ∈ s1 ∧ (a0 = a ∨ ¬ a0 ∈ s0)

User applied rule all left

And obtained the following new goals

Proof step 14:
a ∈ s1, a ∈ s0, a0 ∈ s0 → a0 ∈ s1, a0 = a ∨ a0 ∈ s1 ∧ ¬ a0 ∈ s0 ⊦ # s1 = 0, # s0 = 0, s0 = ∅, s1 = ∅, a0 ∈ s1 ∧ (a0 = a ∨ ¬ a0 ∈ s0)

Rule argument

pos: left 3 terms: a0 abort: F discard: T computed: T

Additional information

Used terms: a0