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Proof step 3 from proof eqpos (previous step)

∀ char1, char2. pos(char1, str0) = pos(char2, str0) ∧ pos(char2, str0) < # str0 → char1 = char2 ∨ str0 = "", pos(char1, char + str0) = pos(char2, char + str0), pos(char2, char + str0) < #(char + str0) ⊦ char1 = char2, char + str0 = ""

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
pos(char1, char ' + str0) = pos(char2, char ' + str0), pos(char2, char ' + str0) ≤ # str0, ∀ char1, char2. pos(char1, str0) = pos(char2, str0) ∧ pos(char2, str0) < # str0 → char1 = char2 ∨ str0 = "" ⊦ char1 = char2

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(char ' + str) = (# str)+1; from specification list[string-new]%a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string-new]%e-01: ⊦ char ' + str ≠ ""; from specification list[string-new]%c: ⊦ char + str = char ' + str; from specification list[string-new]