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Proof step 22 from proof sharing-emp (previous step)

P(H) ⊦ ∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P(H0 ∪ H2) ∧ H1 ∪ H2 = ∅

User applied rule exists right

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ H ⊥ ∅; (s)% ⊦ ∅ ⊥ H; (s)% ⊦ ∅ ∪ H = H; (s)% ⊦ H ∪ ∅ = H; (s) Used terms: H, ∅, ∅