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Proof step 12 from proof put-firstn-restn (previous step)

# y0 = m + 1 ⊦ # y0 = 0, (a0 ' + y0 + z)[# y0, a] = a0 ' + firstn(m, y0 + z) + a ' + z, y0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# y0 = m + 1 ⊦ (y0 + z)[m, a] = firstn(m, y0 + z) + a ' + z, y0 = []

Rule argument

Additional information

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