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Proof step 8 from proof exin-uni (previous step)

⊦ r = null, r ∈ H0, ∃ H1, o0. (H0 ++ r)[r, o] = H1 ∪ (∅ ++ r)[r, o0] ∧ ¬ r ∈ H1

User applied rule exists right

And obtained the following new goals

Proof step 9:
⊦ r = null, r ∈ H0, (H0 ++ r)[r, o] = H0 ∪ (∅ ++ r)[r, o] ∧ ¬ r ∈ H0

Rule argument

pos: right 3 terms: H0, o abort: F discard: T computed: T

Additional information

Used terms: H0, o