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Proof step 6 from proof path-nodups-subset (previous step)

r ∈ H, path(r ' + x0, H), # H ≤ # x0 ⊦ (# H) -1 < # x0 ∧ path(x0, H -- r), dups(x0), r ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
r ∈ H, path(r ' + x0, H), # H ≤ # x0 ⊦ dups(x0), r ∈ x0, path(x0, H -- r)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m ≤ n → (n < m ↔ false)], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0], [CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m], [CSIMPSEQ : ⊦ m < 1 ↔ m = 0], [CSIMPSEQ : ⊦ # H = 0 ↔ H = ∅], [CSIMPSEQ : ⊦ ∅ -- r = ∅], [CSIMPSEQ : ⊦ # ∅ = 0], [CSIMPSEQ : ⊦ 0 ≤ m], [CSIMPSEQ : ⊦ ¬ r ∈ ∅], [CSIMPSEQ : ⊦ n -1 = n - 1])]]