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Proof step 7 from proof btree-ri-unique (previous step)

((btree(rp, null, adr, height, dirty, ri0, iast, as0, alladrs0, ok) * P0))(zh), ((btree(rp, null, adr, height, false, ri1, iast, as1, alladrs1, ok) * P1))(zh), as0 ⊆ alladrs0, as0 ⊆ dom iast, as1 ⊆ alladrs1, as1 ⊆ dom iast, null = null ∨ ¬ zh[null].dirty ⊦ ri0 = ri1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
((btree(rp, null, adr, height, false, ri1, iast, alladrs1, alladrs1, ok) * P1))(zh), ((btree(rp, null, adr, height, dirty, ri0, iast, as0, alladrs0, ok) * P0))(zh), alladrs1 ⊆ dom iast, as0 ⊆ dom iast, as0 ⊆ alladrs0 ⊦ ri0 = ri1

Rule argument

Additional information

Used simplifier rules: btree-clean-alladrs: ⊦ (( btree(rp, r, adr, height, false, ri, iast, as, alladrs, nstat) * P) )(zh) → as = alladrs;%btree-refed-subseteq-alladrs: ⊦ (( btree(rp, r, adr, height, dirty, fi, iast, as, alladrs, nstat) * P) )(zh) → as ⊆ alladrs;%btree-adrs: ⊦ (( btree (rp, rp0, adr0, height, dirty, fi, iast, as, alladrs, nstat) * P) )(zh) → as ⊆ dom iast;% ⊦ as ⊆ as; (s)%sep-com: ⊦ P * P2 = P2 * P; from specification heap-sep[zheap]%sep-ass: ⊦ (P * P3) * P4 = P * P3 * P4; from specification heap-sep[zheap]