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Proof step 3 from proof btree-istore-reduce (previous step)

((btree(rp, r0, adr, height, dirty0, ri, iast0, as, alladrs, nstat) * P))(zh0), iast0 ⊆ iast1, as ⊆ dom iast0, as ⊆ alladrs ⊦ ((btree(rp, r0, adr, height, dirty0, ri, iast1, as, alladrs, nstat) * P))(zh0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ iast0 | as ⊆ iast1 ∧ ((btree(rp, rp0, adr, height, dirty, ri, iast0, as, alladrs, nstat) * P))(zh) → ((btree(rp, rp0, adr, height, dirty, ri, iast1, as, alladrs, nstat) * P))(zh)

Rule argument

spec: <toplevel> name: btree-istore-reduce-refed seq: ⊦ iast0 | as ⊆ iast1 ∧ (( btree (rp, rp0, adr, height, dirty, ri, iast0, as, alladrs, nstat) * P) )(zh) → (( btree (rp, rp0, adr, height, dirty, ri, iast1, as, alladrs, nstat) * P) )(zh) subst: [iast0, zh, P, nstat, alladrs, as, iast1, ri, dirty, height, adr, rp0, rp] → [iast0, zh0, P, nstat, alladrs, as, iast1, ri, dirty0, height, adr, r0, rp] flags: <all paths>

Additional information

Used simplifier rules: subseteq-restrict: ⊦ iast0 ⊆ iast1 → iast0 | as ⊆ iast1; Used terms: iast0, zh0, P, nstat, alladrs, as, iast1, ri, dirty0, height, adr, r0, rp