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Proof step 16 from proof ascsubseq-nodups (previous step)

n ∈ ix1, ascsubseq((m ' + ix) + ix1, x), ∀ n0, ix0, x0. ¬ n0 ∈ ix ∨ ¬ ascsubseq(ix + ix0, x0) ∨ ¬ n0 ∈ ix0 ⊦ n ≠ m ∧ ¬ n ∈ ix

User applied rule constructor cut

And closed the goal!

Rule argument

var:ix

Additional information

[Concretesimprules((Csimpnamedseq(ascsubseq-first-notin-rest,,, ⊦ ascsubseq(n ' + ix, x) → (n ∈ ix ↔ false)) Csimpgen(ilist freely generated by [], +;) Csimpnamedseq(list,indexseq,Nil, ⊦ [] + ix = ix) Csimpnamedseq(list,indexseq,a, ⊦ (ix + ix0) + ix1 = ix + ix0 + ix1) Csimpnamedseq(list,indexseq,in-empty, ⊦ ¬ n ∈ [])))]