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Proof step 100 from proof hasSharing-dd-help (previous step)

# sks = 2 + # refs0 + # refs2, # sks0 = 2 + # refs + # refs1, st[r -- sks] = refval(r0), st[r -- sks0] .refval = ((r ' + refs) + r1 ' + refs1 + r0 ')[2 + # refs + # refs1], validrefs(str, (r ' + refs) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, (r ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), fits(r, sks, st), fits(r, sks0, st), ∀ n. n ≠ 0 ∧ ¬ 2 + # refs + # refs1 < n → st[r -- firstn(n, sks0)] .refval = ((r ' + refs) + r1 ' + refs1 + r0 ')[n], ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = ((r ' + refs0) + r2 ' + refs2 + r0 ')[n] ⊦ refval(r0) = st[r -- sks0], sks = sks0, r2 = r1, r1 = r0, r0 = jvmref, sks0 = [], sks = [], path∃(str, r2, r1, st), path∃(str, r1, r2, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 97:
n ≠ 0 ⊦ (r ' + refs)[(* n)] = refs[n -1]
Proof step 99:
2 + # refs + # refs1 ≠ 0, # sks = 2 + # refs0 + # refs2, # sks0 = 2 + # refs + # refs1, st[r -- sks] = refval(r0), st[r -- sks0] .refval = (refs + r1 ' + refs1 + r0 ')[(2 + # refs + # refs1) -1], validrefs(str, (r ' + refs) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, (r ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), fits(r, sks, st), fits(r, sks0, st), ∀ n. n ≠ 0 ∧ ¬ 2 + # refs + # refs1 < n → st[r -- firstn(n, sks0)] .refval = ((r ' + refs) + r1 ' + refs1 + r0 ')[n], ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = ((r ' + refs0) + r2 ' + refs2 + r0 ')[n] ⊦ refval(r0) = st[r -- sks0], sks = sks0, r2 = r1, r1 = r0, r0 = jvmref, sks0 = [], sks = [], path∃(str, r2, r1, st), path∃(str, r1, r2, st)

Rule argument

[speclemmaarg: n ≠ 0 ,(r ' + refs)[(* n)] = refs[n -1] ,[n, refs, r] → [2 + # refs + # refs1, refs + r1 ' + refs1 + r0 ', r] ,F ,list-del ,references ,get-numrec ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = 0 ↔ m = 0 ∧ n = 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])], [EXTRATERMS : (2 + # refs + # refs1, refs + r1 ' + refs1 + r0 ', r)]]