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Proof step 40 from proof from-append-add (previous step)

(n2 + n1) + n = # x0, # x = n2 + n1, n0 < n2, ∀ n3. n3 < # x0 → x0[n3] = ((σ to n) + ⌜x⌝ )[n3] ⊦ n2 = 0, x = [], n2 + n1 = 0, x0[(n1 + n) + n0] = x[n1 + n0], # σ < ⌜n⌝

User applied rule all left

And obtained the following new goals

Proof step 41:
(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⌝

Rule argument

pos: left 4 terms: (n1 + n) + n0 abort: F discard: T computed: T

Additional information

Used terms: (n1 + n) + n0