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Proof step 5 from proof septract-dual (previous step)

⊦ (∀ H1. H1 ⊥ H ∧ P0(H1) → P1(H1 ∪ H)) ↔ ¬ ((P0 -⊗ ¬ P1))(H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ((P0 -⊗ P1))(H) ↔ (∃ H0, H1. H0 ⊥ H ∧ H1 = H0 ∪ H ∧ P0(H0) ∧ P1(H1))
Proof step 7:
⊦ (∀ H1. H1 ⊥ H ∧ P0(H1) → P1(H1 ∪ H)) ↔ ¬(∃ H0, H1. H0 ⊥ H ∧ H1 = H0 ∪ H ∧ P0(H0) ∧ (¬ P1)(H1))

Rule argument

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

Additional information

Used terms: H, ¬ P1, P0