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Proof step 3 from proof true-elims (previous step)

((P0 ⊗ P))(H) ⊦ ((true ⊗ P))(H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ((P0 ⊗ P1))(H) ↔ (∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P1(H1 ∪ H2))
Proof step 5:
∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P(H1 ∪ H2) ⊦ ((true ⊗ P))(H)

Rule argument

spec: <toplevel> name: sharing seq: ⊦ ((P0 ⊗ P1))(H) ↔ (∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P1(H1 ∪ H2)) subst: [H, P1, P0] → [H, P, P0] flags: <all paths>

Additional information

Used terms: H, P, P0