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Proof step 3 from proof BFS-returns-fail-implies-not-reachable (previous step)

S0 = ∅ ++ source, S1 = ∅ ++ source, S = ∅, n0 = 0, ¬ done0, ∀ n. ¬ reach(source, dest, n), (∃ n. ¬ ¬ reach(source, dest, n)) ∨ ¬ (done0 → dest ∈ S0) ∨ ¬ INV(S0, S, S1, n0, source) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
reach(source, dest, n), ∀ n. ¬ reach(source, dest, n) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ INV (∅ ++ source, ∅, ∅ ++ source, 0, source)])]]