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Proof step 4 from proof adjunction (previous step)

((P0 * P1))(H), ∀ H. P0(H) → ((P1 -* P2))(H) ⊦ P2(H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 6:
∀ H. P0(H) → ((P1 -* P2))(H), ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1) ⊦ P2(H)

Rule argument

spec: <toplevel> name: sep seq: ⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)) subst: [H, P1, P0] → [H, P1, P0] flags: <all paths>

Additional information

Used terms: H, P1, P0