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Proof step 68 from proof ex-validrefs-fits-stronger (conclusion)

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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 67:
validrefs(str, refs, ty, st) ⊦ refs = [], ∃ sks. fits(refs .first, sks, st) ∧ 1 + # sks = # refs ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st[refs .first -- firstn(n, sks)] .refval = refs[n])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ 0 < n ↔ n ≠ 0])]]