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Proof step 5 from proof inv-add1 (previous step)

ro.inodes?, fo.inodes?, log-cons(log, fs), node-cons(nd.key, nd), ri-cons(ri, fs), ri-inj(ri), ri-inj(fi) ⊦ ⟪ choose adr1 with new(adr1 ', fs) ∧ ¬ dups(adr1 ') in {adr := adr1; fs[adr] := nd, log := log + adr, err := ESUCCESS} ifnone err := ENOSPC or {fail#(; ; err); 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

Proof step 6:
fo.inodes?, ro.inodes?, log-cons(log, fs), node-cons(nd.key, nd), ri-cons(ri, fs), ri-inj(fi), ri-inj(ri) ⊦ ⟪choose adr1 with new(adr1 ', fs) ∧ ¬ dups(adr1 ') in {adr := adr1; fs[adr] := nd, log := log + adr, err := ESUCCESS} ifnone err := ENOSPC; boolvar := ?⟫ (log-cons(log, fs) ∧ ri-cons(ri, fs))
Proof step 13:
fo.inodes?, ro.inodes?, log-cons(log, fs), node-cons(nd.key, nd), ri-cons(ri, fs), ri-inj(fi), ri-inj(ri) ⊦ ⟪fail#(; ; err); adr := ?; boolvar := ?⟫ (log-cons(log, fs) ∧ ri-cons(ri, fs))

Proof step 5 from proof inv-add1 (previous step)

Heuristic conditional right split applied rule or split right

And obtained the following new goals

Rule argument

Additional information