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Proof step 14 from proof newref-newreflist-subset (previous step)

∀ r, st, st0. r ∈ newref_list(m, st) ∧ st0 ⊆ st → newref(r, st0) ⊦ r ∈ newref_list(m +1, st) ∧ st0 ⊆ st → newref(r, st0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ 0 < n → newref_list(n, st) = newref(st) + newref_list(n -1, st[newref(st) -- 0 ', noval])
Proof step 3:
⊦ newref_list(n + 1, st) = newref(st) + newref_list(n, st[newref(st) -- 0 ', noval])
Proof step 13:
0 < m +1, ∀ r, st, st0. r ∈ newref_list(m, st) ∧ st0 ⊆ st → newref(r, st0) ⊦ r ∈ newref(st) + newref_list(m +1 -1, st[newref(st) -- 0 ', noval]) ∧ st0 ⊆ st → newref(r, st0)

Rule argument

[crewrite-arg: newrefs-rec-01, F, [st, n] → [st, m +1]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ r + refs = r ' + refs], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ 0 < n ↔ n ≠ 0], [CSIMPSEQ : ⊦ m + 1 ≠ 0])], [EXTRATERMS : (st, m +1)]]