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Proof step 15 from proof wand-shift (previous step)

P0(H1), P1(H), P2(H ∪ H0), H1 ⊥ H0, H ⊥ H1, H ⊥ H0 ⊦ ∃ H3, H2. H3 ⊥ H2 ∧ H ∪ H1 ∪ H0 = H3 ∪ H2 ∧ P0(H3) ∧ P2(H2)

User applied rule exists right

And obtained the following new goals

Proof step 16:
P0(H1), P1(H), P2(H ∪ H0), H ⊥ H0, H ⊥ H1, H1 ⊥ H0 ⊦ H1 ⊥ H ∪ H0 ∧ H ∪ H1 ∪ H0 = H1 ∪ H ∪ H0 ∧ P0(H1) ∧ P2(H ∪ H0)

Rule argument

pos: right 1 terms: H1, H ∪ H0 abort: F discard: T computed: T

Additional information

Used terms: H1, H ∪ H0