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Proof step 1 from proof btree-remove-context (conclusion)

⊦ ((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, ok) * P))(zh) ↔ (∃ P0. P0 ≡ P ∧ ((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, ok) * P0))(zh))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, ok) * P))(zh) ⊦ ∃ P0. P0 ≡ P ∧ ((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, ok) * P0))(zh)
Proof step 7:
∃ P0. P0 ≡ P ∧ ((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, ok) * P0))(zh) ⊦ ((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, ok) * P))(zh)

Rule argument

pos: right 1

Proof step 1 from proof btree-remove-context (conclusion)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information