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Proof step 24 from proof initialization (previous step)

⌊err1⌋ ⊦ ∃ err0. (⌊err0⌋ ∨ err0 = ESUCCESS) ∧ ⟨err := err0; if err = ESUCCESS then avols := ∅ else avols := ?⟩ (avols1 = avols ∧ err1 = err)

User applied rule exists right

And obtained the following new goals

Proof step 25:
⌊err1⌋ ⊦ (⌊err1⌋ ∨ err1 = ESUCCESS) ∧ ⟨err := err1; if err = ESUCCESS then avols := ∅ else avols := ?⟩ (avols1 = avols ∧ err1 = err)

Rule argument

pos: right 1 terms: err1 abort: F discard: T computed: T

Additional information

Used terms: err1