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Proof step 5 from proof inv-add4 (previous step)

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