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Proof step 20 from proof theoremx-atleast-ind (previous step)

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

User applied rule induction

And obtained the following new goals

Proof step 19:
n ∈ ix ∧ inv(ixs + ix ', # x) ∧ precedes∃(ixs + ix ', x), ∀ 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

indvar:# ixs wfpred:<

Additional information

[Concretesimprules((Csimplesspred(<))), Extraterms((# ixs))]