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Proof step 4 from proof inv-add2 (previous step)

jvalid1 ↔ jvalid3, jvalid0 ↔ jvalid4, jvalid ↔ jvalid2, adr = adr1, adr0 = adr2, fi = fi1, fi0 = fi2, jmaxino = jmaxino0, nd = nd1, nd0 = nd2, remove-refsize0(gstore0) = gstore0, elems nx ∩ jro = ∅, dom gstore0 ∪ elems nx = dom gstore ∪ jro, fi1.inos < pmaxino, fi2.inos < jmaxino1, p_os < pmaxino, inv(gstore0), inv(gstore \ jro), refsize-cons(gstore0, range fi1), refsize-cons(gstore, range fi2), ri-inj(fi1), ri-inj(fi2), pmaxino ≤ jmaxino1, fi1.lebs ⊆ dom gstore0, fi2.lebs ⊆ dom gstore, jvalid4 ∧ nx ≠ [] → valid-journal-head(nx, gstore, jro) ⊦ ⟪let ndx = nd0 + nd ', ax = ? in {gjournal_journal_addn#(ndx, jvalid1, fi0, fi, gstore0, pmaxino, p_os; ax, jvalid0, jmaxino, jro, gstore, nx, jvalid; err); if err = ESUCCESS then {adr0 := ax[0]; adr := ax[1]}}⟫ ( (jvalid0 ∧ nx ≠ [] → valid-journal-head(nx, gstore, jro)) ∧ fi2.lebs ⊆ dom gstore ∧ refsize-cons(gstore, range fi2) ∧ inv(gstore \ jro) ∧ elems nx ∩ jro = ∅ ∧ ¬ dups(nx) ∧ dom gstore0 ∪ elems nx = dom gstore ∪ jro), pmaxino = 0, dups(nx)

Heuristic symbolic execution applied rule let right

And obtained the following new goals

Rule argument

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