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Proof step 21 from proof wand-right (previous step)

P(H1), Q0(H0), H1 ⊥ H0, H0 ⊥ H1, ∀ H0. H0 ⊥ H1 ∧ Q0(H0) → Q1(H0 ∪ H1) ⊦ Q1(H0 ∪ H1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
P(H1), Q0(H0), H1 ⊥ H0, ∀ H0. H0 ⊥ H1 ∧ Q0(H0) → Q1(H0 ∪ H1) ⊦ Q1(H0 ∪ H1)

Rule argument

Additional information

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