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Proof step 51 from proof ex-validrefs-fits-stronger (previous step)

0 ≤ i, 0 < st[r -- _length] .intval, is_array(str, r, st), st[r -- _type] .type ≤ ty, i < st[r -- _length] .intval, is_referencevalue(st[r -- i ']), validrefs(str, st[r -- i '] .refval ' + refs1, st[r -- _type] .type .type, st), ∀ st0, ty, r1, str, refs. # refs < # (st[r -- i '] .refval + refs1) → ( validrefs(str, r1 ' + refs, ty, st0) → (∃ sks. fits(r1, sks, st0) ∧ # sks = # refs ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st0[r1 -- firstn(n, sks)] .refval = (r1 ' + refs)[n]))) ⊦ r = jvmref, ∃ sks. fits(r, sks, st) ∧ # sks = 1 + # refs1 ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st[r -- firstn(n, sks)] .refval = (r ' + st[r -- i '] .refval ' + refs1)[n])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 50:
st[r -- i '] = refval(r0), 0 ≤ i, 0 < st[r -- _length] .intval, is_array(str, r, st), st[r -- _type] .type ≤ ty, i < st[r -- _length] .intval, is_referencevalue(st[r -- i ']), validrefs(str, r0 ' + refs1, st[r -- _type] .type .type, st), ∀ st0, ty, r1, str, refs. # refs < # (r0 + refs1) → ( validrefs(str, r1 ' + refs, ty, st0) → (∃ sks. fits(r1, sks, st0) ∧ # sks = # refs ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st0[r1 -- firstn(n, sks)] .refval = (r1 ' + refs)[n]))) ⊦ r = jvmref, ∃ sks. fits(r, sks, st) ∧ # sks = 1 + # refs1 ∧ (∀ n. n ≠ 0 ∧ n ≤ # sks → st[r -- firstn(n, sks)] .refval = (r ' + r0 ' + refs1)[n])

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [st[r -- i '], r0] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (st[r -- i '], r0)]]