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Proof step 10 from proof prefixn-prefixn (previous step)

# y0 = (# y < m ⊃ # y;m), firstn(m, y0 + z0) = y + z, m ≤ # y0 + # z0 ⊦ y = y0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 11:
# y0 = # y, firstn(m, y0 + z0) = y + z, # y < m, m ≤ # y0 + # z0 ⊦ y = y0

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ firstn(# x, x + y) = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ x = x + y ↔ y = []; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)