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Proof step 9 from proof eqpos (previous step)

char2 ≠ char, pos(char1, str0) + 1 = pos(char2, str0) + 1, pos(char1, str0) < # str0, ∀ char0, char3. pos(char0, str0) = pos(char3, str0) ∧ pos(char3, str0) < # str0 → char0 = char3 ⊦ str0 = "", char1 = char, char1 = char2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
pos(char1, str0) = pos(char2, str0), pos(char2, str0) < # str0, ∀ char0, char3. pos(char0, str0) = pos(char3, str0) ∧ pos(char3, str0) < # str0 → char0 = char3 ⊦ char2 = char, char1 = char2, char1 = char, str0 = ""

Rule argument

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2