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

m < # pebs, mtd-cons(pebs) ⊦ ⟪if m < # pebs then {fail_or_success#(; ; err); if err = ESUCCESS then isbad := pebs[m].bad else isbad := ?} else abort⟫ true

Heuristic symbolic execution applied rule if positive right

And obtained the following new goals

Proof step 5:
m < # pebs, mtd-cons(pebs) ⊦ ⟪fail_or_success#(; ; err); if err = ESUCCESS then isbad := pebs[m].bad else isbad := ?⟫ true

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)