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Proof step 11 from proof ex-validrefs-fits (previous step)

validrefs(str, r ' + refs0, ty, st) ⊦ ∃ sks. fits(r, sks, st) ∧ st[r -- sks] .refval = (r ' + refs0) .last

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
validrefs(str, refs, ty, st), refs ≠ [] ⊦ ∃ sks. fits(refs .first, sks, st) ∧ 1 + # sks = # refs ∧ (∀ n. 0 < n ∧ n ≤ # sks → st[refs .first -- firstn(n, sks)] .refval = refs[n])
Proof step 10:
∃ sks. fits((r ' + refs0) .first, sks, st) ∧ 1 + # sks = # (r ' + refs0) ∧ (∀ n. 0 < n ∧ n ≤ # sks → st[(r ' + refs0) .first -- firstn(n, sks)] .refval = (r ' + refs0)[n]), validrefs(str, r ' + refs0, ty, st) ⊦ ∃ sks. fits(r, sks, st) ∧ st[r -- sks] .refval = (r ' + refs0) .last

Rule argument

[LEMMAARG : ex-validrefs-fits-stronger, [str, refs, ty, st] → [str, r ' + refs0, ty, st]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []], [CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])], [EXTRATERMS : (str, r ' + refs0, ty, st)]]