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Proof step 2 from proof at-eq (previous step)

chars[m] ' = chars[n] ', n < # chars, m < # chars ⊦ m = n, chars = [], dups(chars)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ str ≠ "" ∧ str0 ≠ "" → (str = str0 ↔ str.char1 = str0.char1 ∧ str.rstring = str0.rstring)

Rule argument

spec: string-data inst: name: inj-03 seq: ⊦ str ≠ "" ∧ str0 ≠ "" → ( str = str0 ↔ str.char1 = str0.char1 ∧ str.rstring = str0.rstring) subst: [str0, str] → [chars[n] ', chars[m] '] flags: rewrite

Additional information

Used simplifier rules: ⊦ char ' ≠ "";% ⊦ char '.char1 = char;%⊦ m < # chars ∧ n < # chars ∧ ¬ dups(chars) → (chars[m] = chars[n] ↔ m = n); Used terms: chars[n] ', chars[m] '