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Proof step 6 from proof BFS-returns-fail-implies-not-reachable (previous step)

¬ done0, S0 ≠ ∅, # S0 ≠ 0, INV(S0, S, S1, n0, source), v ∈ S0, ∀ n. ¬ reach(source, dest, n), v = dest ⊦ [skip] ((∀ n. ¬ reach(source, dest, n)) ∧ (true → dest ∈ S0) ∧ INV(S0, S, S1, n0, source))

Heuristic symbolic execution applied rule skip right

And obtained the following new goals

Proof step 5:
¬ done0, S0 ≠ ∅, # S0 ≠ 0, INV(S0, S, S1, n0, source), v ∈ S0, ∀ n. ¬ reach(source, dest, n), v = dest, (∃ n. ¬ ¬ reach(source, dest, n)) ∨ ¬ (true → dest ∈ S0) ∨ ¬ INV(S0, S, S1, n0, source) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]