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Proof step 4 from proof eqpos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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: length-one: ⊦ # char ' = 1;% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ #(str + str0) = # str + # str0; (s)%a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string]% ⊦ char ' + str ≠ ""; (s)% ⊦ char + str = char ' + str; (s)