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Proof step 46 from proof theoremx-atmost-h (conclusion)

⊦ ascsubseq(ix, x) ∧ # ixs < # ix ∧ covers(ixs, ix) → (∃ n0, n1, n. n0 ∈ ix ∧ n1 ∈ ix ∧ precedes(n0, n1, x) ∧ n < # ixs ∧ n0 ∈ ixs[n] ∧ n1 ∈ ixs[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
covers(ixs, ix), # ixs < # ix, ascsubseq(ix, x) ⊦ ix = [], ∃ n0, n1, n. n0 ∈ ix ∧ n1 ∈ ix ∧ precedes(n0, n1, x) ∧ n < # ixs ∧ n0 ∈ ixs[n] ∧ n1 ∈ ixs[n]

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(list-data,indexseq,l, ⊦ # ix = 0 ↔ ix = []) Csimpnamedforward(nat-basic1,,f, ⊦ m < n → n ≠ 0)))]