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Proof step 10 from proof ex-validrefs-fits (previous step)

∃ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# sks = # refs0, validrefs(str, r ' + refs0, ty, st), fits(r, sks, st), ∀ n. n ≠ 0 ∧ ¬ # refs0 < n → st[r -- firstn(n, sks)] .refval = (r ' + refs0)[n] ⊦ ∃ sks. fits(r, sks, st) ∧ st[r -- sks] .refval = (r ' + refs0) .last

Rule argument

[emptyarg]

Additional information

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