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Proof step 3 from proof allls-in (previous step)

∀ n, n0. ¬(n ∈ ix0 ∧ all<(ix0, n0) ∧ n0 ≤ n), n ∈ m + ix0, all<(m + ix0, n0), n0 ≤ n ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(allls-rec,,, ⊦ all<(n0 ' + ix, n) ↔ n0 < n ∧ all<(ix, n)) Csimpnamedseq(nat-basic1,,f, ⊦ m < n → n ≠ 0) Csimpnamedforward(nat-basic1,,f, ⊦ m < n → n ≠ 0) Csimpnamedforward(nat,,fle, ⊦ m < n ∧ n ≤ n0 → m < n0) Csimpnamedseq(nat,,le-ls-simp, ⊦ m ≤ n → (n < m ↔ false)) 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)))]