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Proof step 3 from proof btree-remove-context (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 4:
((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, ok) * P))(zh), as ⊆ alladrs, as ⊆ dom iast ⊦ P ≡ P ∧ ((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, ok) * P))(zh)

Rule argument

pos: right 1 terms: P abort: F discard: T computed: T

Additional information

Used terms: P