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Proof step 3 from proof path-nodups-subset (previous step)

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

User applied rule constructor cut

And closed the goal!

Rule argument

[var-arg: x0]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : reflist generated by [], +;], [CSIMPSEQ : ⊦ ¬ r ∈ r0 ' + x0 ↔ r ≠ r0 ∧ ¬ r ∈ x0], [CSIMPSEQ : ⊦ dups(r ' + x0) ↔ dups(x0) ∨ r ∈ x0], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (x0 + x1) = # x0 + # x1], [CSIMPSEQ : ⊦ path(r ' + r0 ' + x, H) ↔ ¬ ¬ ( r ∈ H ∧ r ≠ null ∧ H[r] .nxt = r0 ∧ path(r0 ' + x, H))], [CSIMPSEQ : ⊦ r + x0 = r ' + x0], [CSIMPSEQ : ⊦ x0 + [] = x0], [CSIMPSEQ : ⊦ (x0 + x1) + x2 = x0 + x1 + x2], [CSIMPSEQ : ⊦ path(r ', H) ↔ ¬ ¬ (r ∈ H ∧ r ≠ null)], [CSIMPSEQ : ⊦ # [] = 0], [CSIMPSEQ : ⊦ m ≤ 0 ↔ m = 0], [CSIMPSEQ : ⊦ # H = 0 ↔ H = ∅], [CSIMPSEQ : ⊦ ¬ dups([])], [CSIMPSEQ : ⊦ ¬ r ∈ []], [CSIMPSEQ : ⊦ ¬ path([], H)], [CSIMPSEQ : ⊦ ¬ r ∈ ∅])]]