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Proof step 41 from proof BFS-correct-if-reachable (previous step)

∀ S0, n, n0, S, S1, source, dest, done0. m0 = # S0 ∧ 0 = n - n0 ∧ INVR(S0, S, S1, n0, n, source, dest) ∧ INV(S0, S, S1, n0, source) ∧ (done0 → dest ∈ S0) ∧ (S0 = ∅ → S = ∅) → ⟪while S0 ≠ ∅ ∧ ¬ done0 do { choose v with v ∈ S0 in if v = dest then done0 := true else { S0 := S0 -- v ; ADD(succ(v); S1, S) } ; if S0 = ∅ then S0 := S, S := ∅, n0 := n0 + 1 }⟫ ⟪norfail0 := (done0 ⊃ some(n0); fail)⟫ norfail0 = some(n), ¬ done0, 0 = n - n0, m0 + 1 = # S0, S0 ≠ ∅, INVR(S0, S, S1, n0, n, source, dest), INV(S0, S, S1, n0, source), v ∈ S0, v ≠ dest ⊦ ⟪S0 := S0 -- v⟫ ⟪ADD(succ(v); S1, S)⟫ ⟪if S0 = ∅ then S0 := S, S := ∅, n0 := n0 + 1⟫ ⟪while S0 ≠ ∅ ∧ ¬ done0 do { choose v with v ∈ S0 in if v = dest then done0 := true else { S0 := S0 -- v ; ADD(succ(v); S1, S) } ; if S0 = ∅ then S0 := S, S := ∅, n0 := n0 + 1 }⟫ ⟪norfail0 := (done0 ⊃ some(n0); fail)⟫ norfail0 = some(n)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 40:
∀ S0, n, n0, S, S1, source, dest, done0. m0 = # S0 ∧ 0 = n - n0 ∧ INVR(S0, S, S1, n0, n, source, dest) ∧ INV(S0, S, S1, n0, source) ∧ (done0 → dest ∈ S0) ∧ (S0 = ∅ → S = ∅) → ⟪while S0 ≠ ∅ ∧ ¬ done0 do { choose v with v ∈ S0 in if v = dest then done0 := true else { S0 := S0 -- v ; ADD(succ(v); S1, S) } ; if S0 = ∅ then S0 := S, S := ∅, n0 := n0 + 1 }⟫ ⟪norfail0 := (done0 ⊃ some(n0); fail)⟫ norfail0 = some(n), ¬ done0, 0 = n - n0, m0 + 1 = # S0, S0 ≠ ∅, INVR(S0, S, S1, n0, n, source, dest), INV(S0, S, S1, n0, source), v ∈ S0, v ≠ dest ⊦ ⟪ADD(succ(v); S1, S)⟫ ⟪S0 := S0 -- v ; if S0 = ∅ then S0 := S, S := ∅, n0 := n0 + 1⟫ ⟪while S0 ≠ ∅ ∧ ¬ done0 do { choose v with v ∈ S0 in if v = dest then done0 := true else { S0 := S0 -- v ; ADD(succ(v); S1, S) } ; if S0 = ∅ then S0 := S, S := ∅, n0 := n0 + 1 }⟫ ⟪norfail0 := (done0 ⊃ some(n0); fail)⟫ norfail0 = some(n)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]