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Proof step 4 from proof inv-add5 (previous step)

adr = adr3, adr0 = adr2, adr6 = adr1, adr7 = adr4, adr8 = adr5, nd = nd4, nd0 = nd5, nd2 = nd6, nd3 = nd7, fo.inodes?, ro.inodes?, log-cons(log, fs), node-cons(nd4.key, nd4), node-cons(nd5.key, nd5), node-cons(nd1.key, nd1), node-cons(nd6.key, nd6), node-cons(nd7.key, nd7), ri-cons(ri, fs), ri-inj(fi), ri-inj(ri) ⊦ ⟪ choose adr1, adr2, adr3, adr4, adr5 with new(adr1 ' + adr2 ' + adr3 ' + (adr4 ' + adr5), fs) ∧ ¬ dups(adr1 ' + adr2 ' + adr3 ' + adr4 ' + adr5 ') in {adr8 := adr1, adr7 := adr2, adr6 := adr3, adr0 := adr4, adr := adr5; fs := fs[adr8, nd3][adr7, nd2][adr6, nd1][adr0, nd0][adr, nd], log := ((((log + adr8) + adr7) + adr6) + adr0) + adr, err := ESUCCESS} ifnone err := ENOSPC or {fail#(; ; err); adr8 := ?, adr7 := ?, adr6 := ?, adr0 := ?, adr := ?}; boolvar := ?⟫ (log-cons(log, fs) ∧ ri-cons(ri, fs))

Heuristic conditional right split applied rule or split right

And obtained the following new goals

Rule argument

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