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Proof step 15 from proof sep-ass (previous step)

P0(H2), P1(H), P2(H0), H0 ⊥ H2, H2 ⊥ H, H0 ⊥ H ⊦ ((P0 * P1 * P2))(H2 ∪ H ∪ H0), ((P1 * P2))(H ∪ H0) ∧ H2 ⊥ H ∪ H0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1)
Proof step 17:
P0(H2), P1(H), P2(H0), H0 ⊥ H, H2 ⊥ H, H0 ⊥ H2, H ⊥ H0 ⊦ ((P0 * P1 * P2))(H2 ∪ H ∪ H0), true ∧ H2 ⊥ H ∪ H0

Rule argument

spec: <toplevel> name: split-union seq: ⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1) subst: [H1, H0, P1, P0] → [H0, H, P2, P1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H0, H, P2, P1