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Proof step 9 from proof lastn-restn-01 (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# y = n0 + n1, ∀ x, n0. # x = n0 + n1 → lastn(n1, x) = restn(n0, x) ⊦ lastn(n1, y) + a ' = restn(n0, y + a ')

Rule argument

Additional information

Used simplifier rules: e-02: ⊦ x + a ' ≠ []; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list