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Proof step 42 from proof from-append-add (previous step)

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⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
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, ((σ to n) + ⌜x⌝ )[(n1 + n) + n0] = x[n1 + n0], # σ < ⌜n⌝

Proof step 42 from proof from-append-add (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information