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Proof step 2 from proof eq-one (previous step)

# str = 1, # str0 = 1 ⊦ str = str0 + str1 ↔ str = str0 ∧ str1 = ""

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ # str = 0 ↔ str = ""; from specification list-data[string-new]%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2%len: ⊦ #(str + str0) = # str + # str0; from specification list[string-new]%a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string-new]%app: ⊦ str + "" = str; from specification list[string-new]