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Proof step 2 from proof btree-inv-root-neq-null-fwd (previous step)

btree-inv(rp, adr, istore, istore0, zh, fi, fi0) ⊦ rp = null, rp ∈ zh

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: btree-inv: ⊦ btree-inv(r, adr, istore, istore0, zh, ri, fi) ↔ (∃ height, dirty, as, alladrs. (( btree (null, r, adr, height, dirty, ri, nodes(istore), as, alladrs, ok) * true) )(zh) ∧ refsize-cons(istore, as) ∧ adr ∈ istore ∧ alladrs ⊆ dom nodes(istore)) ∧ (∃ height, dirty, as, alladrs. (( btree (null, null, adr, height, false, fi, nodes(istore0), as, alladrs, ok) * true) )(zh) ∧ refsize-cons(istore0, as) ∧ alladrs ⊆ dom nodes(istore));%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;%btree-root-in-heap: ⊦ r ≠ null ∧ (( btree (rp, r, adr, height, dirty, fi, iast, as, alladrs, nstat) * P) )(zh) → r ∈ zh;%sep-com: ⊦ P * P2 = P2 * P; from specification heap-sep[zheap]%sep-ass: ⊦ (P * P3) * P4 = P * P3 * P4; from specification heap-sep[zheap]