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Proof step 8 from proof lastn-restn-01 (previous step)

# x = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → lastn(n1, x) = restn(n0, x) ⊦ lastn(n1, x.butlast) + x.last ' = restn(n0, x)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
x = y + a ' ∧ x ≠ [], # x = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → lastn(n1, x) = restn(n0, x) ⊦ lastn(n1, y) + a ' = restn(n0, x)

Rule argument

spec: list-last inst: name: last-elim seq: x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a ' subst: [x, a, y] → [x, a, y] flags: []

Additional information

Used simplifier rules: size-01: ⊦ # [] = 0; from specification list-data%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a '; (elim) Used terms: x, a, y