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Proof step 21 from proof theoremx-atleast-ind (conclusion)

⊦ inv(ixs + ix ', # x) ∧ precedes∃(ixs + ix ', x) → (∀ n. n ∈ ix → (∃ ix0. # ix0 = # ixs ∧ ascsubseq(ix0 + n, x)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
n ∈ ix, inv(ixs + ix ', # x), precedes∃(ixs + ix ', x) ⊦ ∃ ix0. # ix0 = # ixs ∧ ascsubseq(ix0 + n ', x)

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(list,indexseq,Last, ⊦ ix + n = ix + n ')))]