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Proof step 11 from proof inv-allocate-ind (previous step)

err0 = ESUCCESS, remove-refsize0(gstore1) = gstore1, remove-refsize0(istore0) = istore0, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, inv(gstore2 \ ginval2), inv(gstore1), inv(istore), inv(istore0), istore0 ⊑ istore, pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2, pstate.index? → pstate.lnum ∈ istore, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪istore[n] := indexblock([], [], 0), pstate := index(n)⟫ ( inv(istore) ∧ istore0 ⊑ istore ∧ (pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2) ∧ (pstate.index? → pstate.lnum ∈ istore)), dups(nx), n ∈ istore

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌊ESUCCESS⌋ ; (s)% ⊦ istore ⊑ istore0 ∧ ¬ n ∈ istore0 → istore ⊑ istore0[n, iblock]; (ss)% ⊦ ¬ index(m).journal?; (s)% ⊦ index(m).index?; (s)% ⊦ index(m).lnum = m; (s)% ⊦ n ∈ istore[n, iblock]; (s)% ⊦ inv(istore) → (inv(istore[n, iblock]) ↔ inv(iblock)); (ss)% ⊦ inv(indexblock([], [], m0)); (s)