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Proof step 4 from proof firstn-prefix-04 (previous step)

# x = (1 + # z) + n0, x = a + y ⊦ x = [], a ' + firstn(n0, y) ⊑ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# y = # z + n0 ⊦ firstn(n0, y) ⊑ y

Rule argument

Additional information

Used simplifier rules: ⊦ x + y ⊑ x + z ↔ y ⊑ z; (s)% ⊦ a ' + x ≠ []; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)