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Proof step 5 from proof inv-commit (previous step)

syn ↔ true, adr = adr1, adr0 = adr2, ginval = ginval2, ginval0 = ginval4, ginval1 = ginval3, gstore = gstore1, gstore0 = gstore2, istore = istore1, istore0 = istore2, maxino = maxino1, maxino0 = maxino2, remove-refsize0(gstore1) = gstore1, remove-refsize0(istore1) = istore1, dom remove-refsize0(gstore2) ∩ ginval4 = ∅, elems nx ∩ ginval4 = ∅, dom gstore1 ∪ elems nx = dom gstore2 ∪ ginval4, inv(gstore2 \ ginval4), inv(gstore1), inv(istore2), inv(istore1), istore1 ⊑ istore2, pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore2, pstate.index? → pstate.lnum ∈ istore2 ⊦ ⟪fail_or_success#(; ; err); if err = ESUCCESS then {istore0 := remove-refsize0(istore0); istore := istore0; gstore0 := remove-refsize0(gstore0); gstore := gstore0; ginval0 := ∅; adr := adr0; ginval := ginval1; nx := []; maxino := maxino0; pstate := unbuffered} else pstate := readonly; ginval1 := ?⟫ ( inv(gstore0 \ ginval0) ∧ inv(gstore) ∧ elems nx ∩ ginval0 = ∅ ∧ ¬ dups(nx) ∧ dom gstore ∪ elems nx = dom gstore0 ∪ ginval0 ∧ remove-refsize0(gstore) = gstore ∧ inv(istore0) ∧ inv(istore) ∧ istore ⊑ istore0 ∧ remove-refsize0(istore) = istore ∧ (pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore0) ∧ (pstate.index? → pstate.lnum ∈ istore0)), dups(nx)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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