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Proof step 7 from proof eqpos (previous step)

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, char ' + str0 = "", char1 = char2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ char1 ≠ char2 → pos(char1, char2 ' + str) = pos(char1, str) + 1
Proof step 9:
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, char ' + str0 = "", char1 = char2

Rule argument

spec: inst: name: Pos-rec seq: ⊦ char1 ≠ char2 → pos(char1, char2 ' + str) = pos(char1, str) + 1 subst: [str, char2, char1] → [str0, char, char2] flags: rewrite, current

Additional information

Used simplifier rules: Pos-found: ⊦ pos(char, char ' + str) = 0;%assoc: ⊦ (str1 + str2) + str3 = str1 + str2 + str3; from specification string-append%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: str0, char, char2