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Proof step 5 from proof wand-same-emp-p (previous step)

P0(H) ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ ((P -* P))(H0) ∧ P0(H1)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: ∅, H abort: F discard: T computed: F

Additional information

Used simplifier rules: wand-same-emp: ⊦ ((P -* P))(∅);% ⊦ ∅ ⊥ H; (s)% ⊦ ∅ ∪ H = H; (s) Used terms: ∅, H