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Proof step 4 from proof n2s-length-06 (previous step)

10 ≤ m, 100000 ≤ m, m < 1000000 ⊦ #(n→s(m / 10) + n→s(m % 10)) = 6, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
10 ≤ m, 100000 ≤ m, m < 1000000 ⊦ m = 0, 5 = # n→s(m / 10)

Rule argument

Additional information

Used simplifier rules: n2s-length-01: ⊦ # n→s(m % 10) = 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-num-04: n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; from specification nat%assoc: ⊦ (str1 + str2) + str3 = str1 + str2 + str3; from specification string-append%len-app: ⊦ #(str + str0) = # str + # str0; from specification string-less