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Proof step 66 from proof ex-validrefs-fits-stronger (previous step)

refs = r + refs0, validrefs(str, refs, ty, st) ⊦ refs = [], ∃ sks. fits(r, sks, st) ∧ 1 + # sks = # refs ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st[r -- firstn(n, sks)] .refval = refs[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
validrefs(str, r ' + refs0, ty, st) ⊦ ∃ sks. fits(r, sks, st) ∧ # sks = # refs0 ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st[r -- firstn(n, sks)] .refval = (r ' + refs0)[n])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]