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Proof step 21 from proof precise-disj (previous step)

H ∪ H5 ∪ H0 = H2 ∪ H3 ∪ H1, P0(H), P0(H3), P1(H2), P1(H5), precise(P1), precise(P0), H2 ⊥ H3, H2 ⊥ H1, H5 ⊥ H0, H3 ⊥ H1, H ⊥ H5, H ⊥ H0 ⊦ H ⊥ H2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ precise(P) ↔ (∀ H, H0, H1, H2. P(H) ∧ P(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0)
Proof step 23:
H ∪ H5 ∪ H0 = H2 ∪ H3 ∪ H1, P0(H3), P0(H), P1(H5), P1(H2), precise(P1), H ⊥ H0, H ⊥ H5, H3 ⊥ H1, H5 ⊥ H0, H2 ⊥ H1, H2 ⊥ H3, ∀ H, H0, H1, H2. P0(H) ∧ P0(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H ⊥ H2

Rule argument

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

Additional information

Used terms: P0