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Proof step 5 from proof inv-set-gblock-refsize (previous step)

ginval = ginval1, ginval0 = ginval2, istore = istore1, istore0 = istore2, maxino = maxino1, maxino0 = maxino2, remove-refsize0(gstore0) = gstore0, remove-refsize0(istore1) = istore1, elems nx ∩ ginval2 = ∅, dom gstore0 ∪ elems nx = dom gstore ∪ ginval2, inv(gstore \ ginval2), inv(gstore0), inv(istore2), inv(istore1), maxino2 ∈ gstore, istore1 ⊑ istore2, pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore, pstate.index? → pstate.lnum ∈ istore2, pstate = unbuffered → syn ⊦ ⟪gstore[maxino0].ref-size := n⟫ ( inv(gstore \ ginval2) ∧ dom gstore0 ∪ elems nx = dom gstore ∪ ginval2 ∧ (pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore) ∧ (pstate.index? → pstate.lnum ∈ istore2) ∧ (pstate = unbuffered → syn)), dups(nx)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ dom gstore[n, gblock] = dom gstore ++ n; (s)% ⊦ n ∈ gstore → dom gstore ++ n = dom gstore; (s)%⊦ inv(gstore \ ns) → (inv(gstore[n, gblock] \ ns) ↔ n ∈ ns ∨ inv(gblock)); (ws)% ⊦ inv(gblock.ref-size:= n0) ↔ inv(gblock); (ss)% ⊦ n ∈ gstore ∧ ¬ n ∈ ns ∧ inv(gstore \ ns) → inv(gstore[n]); (ss)% ⊦ ¬ unbuffered.journal?; (s)% ⊦ n ∈ gstore[n0, gblock] ↔ n = n0 ∨ n ∈ gstore; (ws)