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Proof step 3 from proof and-combine (previous step)

P0(H0), P1(H0), precise(P0), H ⊥ H0 ⊦ ((P * (P0 ∧ P1)))(H ∪ H0) ↔ P(H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ P0 * P1 = P1 * P0
Proof step 5:
P0(H0), P1(H0), precise(P0), H ⊥ H0 ⊦ (((P0 ∧ P1) * P))(H ∪ H0) ↔ P(H)

Rule argument

spec: <toplevel> name: sep-com seq: ⊦ P0 * P1 = P1 * P0 subst: [P1, P0] → [P0 ∧ P1, P] flags: <all paths>

Additional information

Used terms: P0 ∧ P1, P