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Proof step 7 from proof crash-neutral-recover (previous step)

fi0 = fi, fo0 = fo, fs0 = fs, log0 = log, ro.inodes?, fo.inodes?, log-cons(log, fs), ri-inj(ri), ri-inj(fi) ⊦ (⌊ESUCCESS⌋ ∨ ESUCCESS = ESUCCESS) ∧ ⟨err := ESUCCESS; if err = ESUCCESS then {ri := fi, ro := fo, ax := log; ks := ro; boolvar := true} else ri := ?, fi := ?, fs := ?, log := ?, ro := ?, fo := ?, ax := ?, ks := ?, boolvar := ?⟩ (boolvar ∧ fi0 = fi ∧ fo0 = fo ∧ fs0 = fs ∧ log0 = log), ∃ err0. (⌊err0⌋ ∨ err0 = ESUCCESS) ∧ ⟨err := err0; if err = ESUCCESS then {ri := fi, ro := fo, ax := log; ks := ro; boolvar := true} else ri := ?, fi := ?, fs := ?, log := ?, ro := ?, fo := ?, ax := ?, ks := ?, boolvar := ?⟩ (boolvar ∧ fi0 = fi ∧ fo0 = fo ∧ fs0 = fs ∧ log0 = log)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
fi0 = fi, fo0 = fo, fs0 = fs, log0 = log, fo.inodes?, ro.inodes?, log-cons(log, fs), ri-inj(fi), ri-inj(ri) ⊦ ⟨err := ESUCCESS; if err = ESUCCESS then {ri := fi, ro := fo, ax := log; ks := ro; boolvar := true} else ri := ?, fi := ?, fs := ?, log := ?, ro := ?, fo := ?, ax := ?, ks := ?, boolvar := ?⟩ (boolvar ∧ fi0 = fi ∧ fo0 = fo ∧ fs0 = fs ∧ log0 = log), ∃ err0. ⌊err0⌋ ∧ ⟨err := err0; if err = ESUCCESS then {ri := fi, ro := fo, ax := log; ks := ro; boolvar := true} else ri := ?, fi := ?, fs := ?, log := ?, ro := ?, fo := ?, ax := ?, ks := ?, boolvar := ?⟩ (boolvar ∧ fi0 = fi ∧ fo0 = fo ∧ fs0 = fs ∧ log0 = log)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ ⌊ESUCCESS⌋ ; (s)