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Proof step 41 from proof ex-validrefs-fits-stronger (previous step)

st[r -- i '] = refval(r0), # sks = # refs1, 0 ≤ i, 0 < st[r -- _length] .intval, is_array(str, r, st), st[r -- _type] .type ≤ ty, i < st[r -- _length] .intval, validrefs(str, r0 ' + refs1, st[r -- _type] .type .type, st), fits(r0, sks, st), n ≤ 1 + # refs1, ∀ n. n ≠ 0 ∧ ¬ # refs1 < n → st[r0 -- firstn(n, sks)] .refval = (r0 ' + refs1)[n], ∀ 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]))) ⊦ n = 0, r = jvmref, st[r -- (i ' ' + firstn(n - 1, sks))] .refval = (r ' + r0 ' + refs1)[n], is_classtype(st[r -- _type] .type)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 40:
n = m + 1, st[r -- i '] = refval(r0), # sks = # refs1, 0 ≤ i, 0 < st[r -- _length] .intval, is_array(str, r, st), st[r -- _type] .type ≤ ty, i < st[r -- _length] .intval, validrefs(str, r0 ' + refs1, st[r -- _type] .type .type, st), fits(r0, sks, st), n ≤ 1 + # refs1, ∀ n. n ≠ 0 ∧ ¬ # refs1 < n → st[r0 -- firstn(n, sks)] .refval = (r0 ' + refs1)[n], ∀ 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]))) ⊦ n = 0, r = jvmref, st[r -- (i ' ' + firstn(m, sks))] .refval = (r ' + r0 ' + refs1)[n], is_classtype(st[r -- _type] .type)

Rule argument

[speclemmaarg: n ≠ 0 ,m = n - 1 ↔ n = m + 1 ,[n, m] → [n, m] ,T ,nat , ,elim-pred-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (n, m)]]