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Proof step 18 from proof theoremx-atleast-ind (previous step)

precedes∃(ixs + ix ', x), inv(ixs + ix ', # x), n ∈ ix, ∀ ixs0, n, ix, x. # ixs0 < # ixs → n ∈ ix ∧ inv(ixs0 + ix ', # x) ∧ precedes∃(ixs0 + ix ', x) → (∃ ix0. # ix0 = # ixs0 ∧ ascsubseq(ix0 + n ', x)) ⊦ ∃ ix0. # ix0 = # ixs ∧ ascsubseq(ix0 + n ', x)

User applied rule cut formula

And obtained the following new goals

Proof step 6:
ixs = [], precedes∃(ixs + ix ', x), inv(ixs + ix ', # x), n ∈ ix, ∀ ixs0, n, ix, x. # ixs0 < # ixs → n ∈ ix ∧ inv(ixs0 + ix ', # x) ∧ precedes∃(ixs0 + ix ', x) → (∃ ix0. # ix0 = # ixs0 ∧ ascsubseq(ix0 + n ', x)) ⊦ ∃ ix0. # ix0 = # ixs ∧ ascsubseq(ix0 + n ', x)

Rule argument

fma:ixs = []

Additional information

[Concretesimprules(()), Extrafmas((ixs = []))]