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Proof step 6 from proof precise-disj (previous step)

H1 ∪ H3 = H ∪ H4, ((P0 * P1 * true))(H ∪ H4), ((P1 * true))(H4), P0(H), P1(H1), precise(P0), precise(P1), H1 ⊥ H3, H ⊥ H4 ⊦ H ⊥ H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ P0 * P1 = P1 * P0
Proof step 8:
H1 ∪ H3 = H ∪ H4, ((P1 * true))(H4), (((P1 * true) * P0))(H ∪ H4), P0(H), P1(H1), precise(P1), precise(P0), H ⊥ H4, H1 ⊥ H3 ⊦ H ⊥ H1

Rule argument

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

Additional information

Used terms: P1 * true, P0