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Proof step 2 from proof indef (previous step)
| a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 |
User applied rule exists right
And closed the goal!
Rule argument
pos: right 1 terms: s -- a abort: F discard: F computed: FAdditional information
Used simplifier rules:
del-addeq-02: ⊦ a ∈ s → (s -- a) ++ a = s;%ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);
Used terms: s -- a