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Proof step 6 from proof eqpos (previous step)

char1 ≠ char, pos(char1, str0) + 1 = 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
pos(char1, str0) + 1 = pos(char2, char ' + str0), pos(char1, str0) < # str0, ∀ char0, char3. pos(char0, str0) = pos(char3, str0) ∧ pos(char3, str0) < # str0 → char0 = char3 ⊦ str0 = "", char1 = char, char1 = char2

Rule argument

Additional information

Used simplifier rules: l: ⊦ # str = 0 ↔ str = ""; from specification list-data[string-new]%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat