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Proof step 3 from proof prop-sep (previous step)

⊦ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ ((⌜bv⌝ ∧ P0))(H0) ∧ P1(H1)) ↔ bv ∧ ((P0 * P1))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
((P0 * P1))(H) ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)

Rule argument

Additional information

Used simplifier rules: and: ⊦ ((P0 ∧ P1))(H) ↔ P0(H) ∧ P1(H);%prop: ⊦ ⌜bv⌝ (H) ↔ bv;%and-ass: ⊦ ((P0 ∧ P1) ∧ P2) = (P0 ∧ P1 ∧ P2);%and-com: ⊦ (P0 ∧ P1) = (P1 ∧ P0);%split-union: ⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1);%sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2;%sep-com: ⊦ P0 * P1 = P1 * P0;