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Proof step 50 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), m0 +1 = # 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 49:
∀ 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), 0 = n - n0, m0 + 1 = # S0, S0 ≠ ∅, INVR(S0, S, S1, n0, n, source, dest), INV(S0, S, S1, n0, source), done0 → dest ∈ S0 ⊦ ⟪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 : ([CSIMPFORWARD : ⊦ s ≠ ∅ → # s ≠ 0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ # ∅ = 0], [CSIMPSEQ : ⊦ m +1 = m + 1])]]