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Proof step 9 from proof path-nodups-subset (previous step)

path(r ' + x0, H), # H ≤ # x0, (∃ H0. # H0 < # x0 ∧ path(x0, H0)) → x0 = [] ⊦ dups(x0), r ∈ x0

User applied rule case distinction

And obtained the following new goals

Proof step 8:
path(r ' + x0, H), # H ≤ # x0 ⊦ ∃ H0. # H0 < # x0 ∧ path(x0, H0), dups(x0), r ∈ x0

Rule argument

[CASEDARG : pos: left 3, ()]

Additional information

[[CONCRETESIMPRULES : ([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 : ⊦ ¬ r ∈ ∅])]]