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Proof step 5 from proof eqpos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ char ≠ char0 → pos(char, char0 ' + str) = pos(char, str) + 1
Proof step 7:
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, char1 = char

Rule argument

spec: list-del, inst: string name: pos-n seq: ⊦ char ≠ char0 → pos(char, char0 ' + str) = pos(char, str) + 1 subst: [str, char0, char] → [str0, char, char1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ pos(char, char ' + str) = 0; (s)%a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string]% ⊦ char ≠ char0 → pos(char, char0 ' + str) = pos(char, str) + 1; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: str0, char, char1