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Proof step 5 from proof precise-same (previous step)

H0 ∪ H = H1 ∪ H2, P(H1), P(H0), precise(P), H1 ⊥ H2, H0 ⊥ H ⊦ H0 = H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ precise(P) ↔ (∀ H, H0, H1, H2. P(H) ∧ P(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0)
Proof step 7:
H0 ∪ H = H1 ∪ H2, P(H0), P(H1), H0 ⊥ H, H1 ⊥ H2, ∀ H, H0, H1, H2. P(H) ∧ P(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H0 = H1

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] → [P] flags: <all paths>

Additional information

Used terms: P