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Proof step 1 from proof eqpos (conclusion)

⊦ pos(char1, str) < # str → (pos(char1, str) = pos(char2, str) ↔ char1 = char2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
pos(char1, str) = pos(char2, str), pos(char2, str) < # str ⊦ char1 = char2, str = ""

Rule argument

Additional information

Used simplifier rules: len-zero: ⊦ # str = 0 ↔ str = "";%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1