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Proof step 8 from proof btree-ri-unique (previous step)

((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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ r = null ∨ ¬ zh0[r].dirty → ( ((btree(rp, r, adr, height, dirty, ri, iast, as, alladrs, nstat) * P))(zh0) ↔ ((btree(rp, r, adr, height, false, ri, iast, as, alladrs, nstat) * P))(zh0))

Rule argument

spec: <toplevel> name: btree-null-or-clean-below seq: ⊦ r = null ∨ ¬ zh0[r].dirty → ( (( btree (rp, r, adr, height, dirty, ri, iast, as, alladrs, nstat) * P) )(zh0) ↔ (( btree (rp, r, adr, height, false, ri, iast, as, alladrs, nstat) * P) )(zh0)) subst: [zh0, P, nstat, alladrs, as, iast, ri, dirty, height, adr, r, rp] → [zh, P0, ok, alladrs0, as0, iast, ri0, dirty, height, adr, null, rp] flags: <all paths>

Additional information

Used simplifier rules: btree-clean-alladrs: ⊦ (( btree(rp, r, adr, height, false, ri, iast, as, alladrs, nstat) * P) )(zh) → as = alladrs;%btree-clean-fwd: ⊦ (( btree (rp0, r0, adr, height0, false, ri0, iast, as0, alladrs0, nstat) * P0) )(zh0) ∧ (( btree (rp1, r1, adr, height1, false, ri1, iast, as1, alladrs1, nstat) * P1) )(zh1) → height0 = height1 ∧ as0 = as1 ∧ ri0 = ri1 ∧ alladrs0 = alladrs1;%sep-ass: ⊦ (P * P3) * P4 = P * P3 * P4; from specification heap-sep[zheap]%sep-com: ⊦ P * P2 = P2 * P; from specification heap-sep[zheap] Used terms: zh, P0, ok, alladrs0, as0, iast, ri0, dirty, height, adr, null, rp