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Proof step 6 from proof inv-sync (previous step)

ginval = ginval1, ginval0 = ginval2, gstore = gstore1, gstore0 = gstore2, istore = istore1, istore0 = istore2, pstate = readonly, remove-refsize0(istore1) = istore1, remove-refsize0(gstore1) = gstore1, elems nx ∩ ginval2 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval2, btree-inv(rp, adr, istore2, istore1, zh, ri, fi), inv(istore1), inv(istore2), inv(gstore1), inv(gstore2 \ ginval2), istore1 ⊑ istore2, pstate = unbuffered → syn, pstate.index? → pstate.lnum ∈ istore2, pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2 ⊦ ⟪err := EROFS⟫ ( (pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2) ∧ (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: ⊦ unbuffered ≠ readonly; (s)% ⊦ ¬ readonly.index?; (s)% ⊦ ¬ readonly.journal?; (s)