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Proof step 6 from proof sublist-last (previous step)

m0 + n0 < # y, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).first ' + firstn(n0, restn(m0, x).rest)).last = x[m0 + n0] ∨ x = [] ⊦ (restn(m0, a ' + y).rest.first ' + firstn(n0, restn(m0, a ' + y).rest.rest)).last = y[m0 + n0], y = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ x.rest = restn(1, x)
Proof step 8:
m0 + n0 < # y, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).first ' + firstn(n0, restn(m0, x).rest)).last = x[m0 + n0] ∨ x = [] ⊦ (restn(1, restn(m0, a ' + y)).first ' + firstn(n0, restn(1, restn(m0, a ' + y)).rest)).last = y[m0 + n0], y = []

Rule argument

spec: inst: name: rest-rest1 seq: ⊦ x.rest = restn(1, x) subst: [x] → [restn(m0, a ' + y)] flags: rewrite, current

Additional information

Used terms: restn(m0, a ' + y)