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Proof step 17 from proof ascsubseq-nodups (previous step)

∀ n, ix1, x. ¬(n ∈ ix ∧ ascsubseq(ix + ix1, x) ∧ n ∈ ix1), n ∈ m + ix, ascsubseq((m + ix) + ix1, x), n ∈ ix1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
n ∈ ix1, ascsubseq((m ' + ix) + ix1, x), ∀ n0, ix0, x0. ¬ n0 ∈ ix ∨ ¬ ascsubseq(ix + ix0, x0) ∨ ¬ n0 ∈ ix0 ⊦ n ≠ m ∧ ¬ n ∈ ix

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(list,indexseq,a, ⊦ (ix + ix0) + ix1 = ix + ix0 + ix1) Csimpnamedseq(list,indexseq,in-17, ⊦ n ∈ n0 ' ↔ n = n0) Csimpnamedseq(list,indexseq,in-append, ⊦ n ∈ ix + ix0 ↔ ¬ ¬(n ∈ ix ∨ n ∈ ix0)) Csimpnamedseq(list,indexseq,c, ⊦ n + ix = n ' + ix)))]