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Proof step 6 from proof pos-range (previous step)

char ≠ char0, # str0 + 1 < pos(char, str0) + 1, ∀ char. # str0 < pos(char, str0) → pos(char, str0) = 0 ⊦ pos(char, str0) + 1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# str0 < pos(char, str0), ∀ char. # str0 < pos(char, str0) → pos(char, str0) = 0 ⊦ char = char0, pos(char, str0) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2