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Proof step 8 from proof mapex-new (previous step)

P(H) ⊦ r = null, r ∈ H, ∃ H1. ¬ r ∈ H1 ∧ H ++ r = (∅ ++ r)[r, (H ++ r)[r]] ∪ H1 ∧ P(H1)

User applied rule exists right

And obtained the following new goals

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

Rule argument

pos: right 3 terms: H abort: F discard: T computed: F

Additional information

Used terms: H