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Proof step 11 from proof del (previous step)

∀ a1. a1 = a ∨ a1 ∈ s → (∃ s1. s ++ a = s1 ++ a1 ∧ ¬ a1 ∈ s1) ⊦ a0 = a, a ∈ s, a0 ∈ s, ∃ s1. s ++ a ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: s ++ a0 abort: F discard: F computed: F

Additional information

Used simplifier rules: eq-04: ⊦ s ++ b ++ a = s ++ a ++ b ↔ true;%in-02: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s); Used terms: s ++ a0