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Proof step 10 from proof diffgetn-nodupssize (previous step)

n ≤ # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ x0 = [], #(x0 \ (a ' + x0)[n] ') + 1 = # x0, a = (a ' + x0)[n], dups(x0), a ∈ x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ ¬ n < # x → (x + y)[n] = y[n - # x]
Proof step 12:
¬ n < # a ', n ≤ # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ x0 = [], #(x0 \ x0[n - # a '] ') + 1 = # x0, a = x0[n - # a '], dups(x0), a ∈ x0

Rule argument

spec: list-del inst: name: get-append-grt seq: ⊦ ¬ n < # x → (x + y)[n] = y[n - # x] subst: [n, y, x] → [n, x0, a '] flags: rewrite

Additional information

Used simplifier rules: get-append-less: ⊦ n < # x → (x + y)[n] = x[n]; from specification list-del%a: ⊦ (x + y) + z = x + y + z; from specification list%len-02: ⊦ # a ' = 0 +1; from specification list%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%one-03: ⊦ a '[0] = a; from specification list-del Used terms: n, x0, a '