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Proof step 36 from proof pushx-covers (previous step)

n < # x, covers(ixs, n) ⊦ x = [], covers(pushx(n, ixs, x), n + 1)

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ n < # x ∧ ixs = pushx(n, ixs0, x) → ixs = ixs0 + n ' ' ∨ (∃ m. m < # ixs0 ∧ ixs = ixs0[m, n + ixs0[m]])

Rule argument

spec: inst: name:pushx-nonrec seq: ⊦ n < # x ∧ ixs = pushx(n, ixs0, x) → ixs = ixs0 + n ' ' ∨ (∃ m. m < # ixs0 ∧ ixs = ixs0[m, n + ixs0[m]]) subst:[n, x, ixs, ixs0] → [n, x, pushx(n, ixs, x), ixs] flags:current, []

Additional information

[Concretesimprules(()), Extraterms((nxpushx(n, ixs, x)ixs))]