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Proof step 28 from proof BFS-correct-if-reachable (previous step)

⟪while S6 ≠ ∅ ∧ ¬ done1 do { choose v with v ∈ S6 in if v = dest then done1 := true else { S6 := S6 -- v ; ADD(succ(v); S4, S5) } ; if S6 = ∅ then S6 := S5, S5 := ∅, n0 := n0 + 1 }⟫ ⟪norfail0 := (done1 ⊃ some(n0); fail)⟫ norfail0 = some(n), ⟪ADD(succ(v); S1, S)⟫ (INV(S0 -- v, S, S1, n0, source) ∧ INVR(S0 -- v, S, S1, n0, n, source, dest)), 〈ADD(succ(v); S1, S)〉 (S1 = s ∧ S = s0), m0 + 1 = # S0, 0 = n - n0, ¬ done, S2 = S0 -- v, S0 ≠ ∅ ++ v, v ≠ dest, S0 ≠ ∅, INVR(S0 -- v, s0, s, n0, n, source, dest), INV(S0 -- v, s0, s, n0, source), v ∈ S0, INV(S0, S, S1, n0, source), INVR(S0, S, S1, n0, n, source, dest), (done1 ↔ false) ∧ S4 = s ∧ S5 = s0 ∧ S6 = S0 -- v ⊦ ⟪while S2 ≠ ∅ ∧ ¬ done do { choose v with v ∈ S2 in if v = dest then done := true else { S2 := S2 -- v ; ADD(succ(v); s, s0) } ; if S2 = ∅ then S2 := s0, s0 := ∅, n0 := n0 + 1 }⟫ ⟪norfail0 := (done ⊃ some(n0); fail)⟫ norfail0 = some(n)

Proof step 28 from proof BFS-correct-if-reachable (previous step)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

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