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

# str = 1, 1 < # str0 ⊦ 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%one-02: ⊦ m ≠ 1 → (m + n = 1 ↔ m = 0 ∧ n = 1); from specification nat-basic2%l: ⊦ # str = 0 ↔ str = ""; from specification list-data[string-new]%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string-new]%len: ⊦ #(str + str0) = # str + # str0; from specification list[string-new]