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Proof step 4 from proof lastN-append-01 (previous step)

n ≤ # y, ∀ n, y. n ≤ # y → lastn(n, x0 + y) = lastn(n, y) ⊦ lastn(n, (a ' + x0) + y) = lastn(n, y)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n < # x → lastn(n, x) = lastn(n, x.rest)
Proof step 6:
n < #((a ' + x0) + y), n ≤ # y, ∀ n, y. n ≤ # y → lastn(n, x0 + y) = lastn(n, y) ⊦ lastn(n, ((a ' + x0) + y).rest) = lastn(n, y)

Rule argument

spec: list-del inst: name: lastN-rec-02 seq: ⊦ n < # x → lastn(n, x) = lastn(n, x.rest) subst: [x, n] → [(a ' + x0) + y, n] flags: rewrite, current

Additional information

Used simplifier rules: len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%l-02: ⊦ ¬ m + n < m; from specification nat-basic2 Used terms: (a ' + x0) + y, n