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

P0(H1), Q0(H1), Q0 ⊆ P0, H1 ⊥ H0, ∀ H1. H1 ⊥ H0 ∧ P0(H1) → P1(H1 ∪ H0) ⊦ P1(H1 ∪ H0), Q1(H1 ∪ H0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: H1

Additional information

Used terms: H1