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Proof step 23 from proof ordered-precedes (previous step)

ordered(ix, x), n0 ∈ ix, n1 ∈ ix, precedes(n0, n1, x) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ n ∈ ix ↔ (∃ ix0, ix1. ix = ix0 + n ' + ix1 ∧ ¬ n ∈ ix0)

Rule argument

spec:list inst:indexseq name:firstoc-rew seq: ⊦ n ∈ ix ↔ (∃ ix0, ix1. ix = ix0 + n ' + ix1 ∧ ¬ n ∈ ix0) subst:[ix, n] → [ix, n0] flags:rewrite

Additional information

[Concretesimprules(()), Extraterms((ixn0))]