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Proof step 13 from proof precise-disj (previous step)

H ∪ H4 = H2 ∪ H0, ((true * P0))(H0), ((P1 * true))(H4), P0(H), P1(H2), precise(P0), precise(P1), H2 ⊥ H0, H ⊥ H4 ⊦ H ⊥ H2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ P0 * P1 = P1 * P0
Proof step 15:
H ∪ H4 = H2 ∪ H0, ((P1 * true))(H4), ((P0 * true))(H0), P0(H), P1(H2), precise(P1), precise(P0), H ⊥ H4, H2 ⊥ H0 ⊦ H ⊥ H2

Rule argument

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

Additional information

Used terms: P0, true