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Proof step 9 from proof del (previous step)

a ∈ s0, ∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1) ⊦ a0 ∈ s0, ∃ s1. s0 ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

User applied rule all left

And obtained the following new goals

Proof step 10:
a ∈ s0, a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1), ∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1) ⊦ a0 ∈ s0, ∃ s1. s0 ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

Rule argument

pos: left 2 terms: a abort: F discard: F computed: T

Additional information

Used terms: a