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Proof step 37 from proof precise-disj (previous step)

H0 ∪ H2 = H1 ∪ H3, (P0 * P1 * true)(H1 ∪ H3), H0 ⊥ H2, H1 ⊥ H3, P0(H0), P1(H1), precise(P0), precise(P1) ⊦ H0 ⊥ H1

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ (P0 * P1)(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: sep antecedent: succedent: (P0 * P1)(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)) substlist: [P0, P1, H] → [P0, P1 * true, H1 ∪ H3] add preconditions?: F is rewrite lemma?: F current spec?: T rotate lemma?: F replace all?: F paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (P0, P1 * true, H1 ∪ H3)]]