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Proof step 2 from proof len-app (previous step)

∀ str0. #(str1 + str0) = # str1 + # str0 ⊦ #((char + str1) + str0) = #(char + str1) + # str0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ str0. #(str1 + str0) = # str1 + # str0 ⊦ #(str1 + str0) = # str1 + # str0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(char ' + str) = (# str)+1; from specification list[string-new]%a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string-new]%c: ⊦ char + str = char ' + str; from specification list[string-new]