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Proof step 12 from proof lastn-restn-01 (previous step)

# y = n0 + n1, n0 ≤ # y, ∀ x, n0. # x = n0 + n1 → lastn(n1, x) = restn(n0, x) ⊦ lastn(n1, y) + a ' = restn(n0, y) + a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# y = n0 + n1, ∀ x, n0. # x = n0 + n1 → lastn(n1, x) = restn(n0, x) ⊦ lastn(n1, y) = restn(n0, y)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = z + y ↔ x = z; (s)