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Proof step 5 from proof pos-get (previous step)

n ≤ # x0, ∀ n0. n0 < # x0 → pos(x0[n0], x0) = n0 ∨ x0 = [] ⊦ pos((a ' + x0)[n], a ' + x0) = n, dups(x0), a ∈ x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ¬ n < # x → (x + y)[n] = y[n - # x]
Proof step 7:
n ≤ # x0, ∀ n0. n0 < # x0 → pos(x0[n0], x0) = n0 ∨ x0 = [] ⊦ pos(x0[n - # a '], a ' + x0) = n, n < # a ', dups(x0), a ∈ x0

Rule argument

spec: <toplevel> name: get-append-grt seq: ⊦ ¬ n < # x → (x + y)[n] = y[n - # x] subst: [n, y, x] → [n, x0, a '] flags: <all paths>

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;%pos-y: ⊦ pos(a, a ' + x) = 0;% ⊦ # a ' = 0 +1; (s)% ⊦ m < 1 ↔ m = 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n, x0, a '