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

⊦ ⟪fail_or_success#(; ; err); if err = ESUCCESS then ri := ∅, fs := ∅, fi := ∅, fo := ∅, ro := ∅, log := [], boolvar0 := true else ri := ?, fs := ?, fi := ?, fo := ?, ro := ?, log := ?, boolvar0 := ?⟫ (err = ESUCCESS → boolvar0 ∧ ri-inj(ri) ∧ ri-inj(fi) ∧ log-cons(log, fs) ∧ ro.inodes? ∧ fo.inodes?)

Heuristic unfold applied rule call right

And obtained the following new goals

Proof step 5:
⊦ ⟪choose err0 with ⌊err0⌋ ∨ err0 = ESUCCESS in err := err0; if err = ESUCCESS then ri := ∅, fs := ∅, fi := ∅, fo := ∅, ro := ∅, log := [], boolvar0 := true else ri := ?, fs := ?, fi := ?, fo := ?, ro := ?, log := ?, boolvar0 := ?⟫ (err = ESUCCESS → boolvar0 ∧ ri-inj(ri) ∧ ri-inj(fi) ∧ log-cons(log, fs) ∧ ro.inodes? ∧ fo.inodes?)

Rule argument

pos: right 1

Additional information

Used simplifier rules: fail_or_success#(; ; ERR) {choose ERR0 with ⌊ERR0⌋ ∨ ERR0 = ESUCCESS in ERR := ERR0};