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Proof step 18 from proof wand-elim (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: H0

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H0