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Proof step 29 from proof recovery (previous step)

err1 = ESUCCESS, fi = fi0, ginval = ginval2, ginval0 = ginval3, gstore = gstore1, gstore0 = crash(gstore4, gstore1, ginval3, nx1), maxino = maxino1, nx = nx1, remove-refsize0(gstore1) = gstore1, remove-refsize0(istore1) = istore1, elems nx1 ∩ ginval3 = ∅, dom gstore1 ∪ elems nx1 = dom gstore4 ∪ ginval3, btree-inv(rp1, adr2, istore2, istore1, zh, fi1, fi0), btree-inv(rp1, adr2, istore2, istore1, zh, ri, fi3), inv(gstore4 \ ginval3), inv(gstore1), inv(istore2), inv(istore1), inv(crash(gstore4, gstore1, ginval3, nx1) \ ginval3), ri-inj(fi1), ri-inj(fi0), istore1 ⊑ istore2, pstate1.journal? → nx1 ≠ [] ∧ pstate1.lnum = nx1.last ∧ pstate1.lnum ∈ gstore4, pstate1.index? → pstate1.lnum ∈ istore2 ⊦ ⟨maxino0 := maxino, syn := true, nx0 := nx, ginval1 := ginval, fi0 := fi⟩ ( ginval0 = ginval3 ∧ gstore0 = crash(gstore4, gstore1, ginval3, nx0) ∧ gstore = gstore1 ∧ nx = nx0 ∧ maxino = maxino0 ∧ ginval = ginval1 ∧ btree-inv(null, adr2, remove-refsize0(crash(istore2, istore1)), istore1, zh0, fi0, fi) ∧ syn ∧ maxino1 = maxino0 ∧ nx1 = nx0 ∧ ginval2 = ginval1 ∧ ESUCCESS = err1), dups(nx1)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ istore0 ⊑ istore ∧ remove-refsize0(istore0) = istore0 ∧ btree-inv(rp, adr, istore, istore0, zh0, ri, fi) ∧ inv(istore) → btree-inv (null, adr, remove-refsize0(crash(istore, istore0)), istore0, zh1, fi, fi); (ss)