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Proof step 41 from proof from-append-add (previous step)

(n2 + n1) + n = # x0, # x = n2 + n1, n0 < n2, (n1 + n) + n0 < # x0 → x0[(n1 + n) + n0] = ((σ to n) + ⌜x⌝ )[(n1 + n) + n0] ⊦ n2 = 0, x = [], n2 + n1 = 0, x0[(n1 + n) + n0] = x[n1 + n0], # σ < ⌜n⌝

User applied rule case distinction

And obtained the following new goals

Proof step 42:
x0[(n1 + n) + n0] = ((σ to n) + ⌜x⌝ )[(n1 + n) + n0], (n2 + n1) + n = # x0, # x = n2 + n1, n0 < n2 ⊦ n2 = 0, x = [], n2 + n1 = 0, x0[(n1 + n) + n0] = x[n1 + n0], # σ < ⌜n⌝

Rule argument

pos: left 4

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l-06: ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; from specification nat-basic2