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Proof step 7 from proof append-eqlen-01 (previous step)

# str0 = # str2, (char ' + str3) + str0 = str1 + str2, ∀ str0, str2, str1. # str0 = # str2 ∧ str3 + str0 = str1 + str2 → str3 = str1 ∧ str0 = str2 ⊦ str1 = "", char = str1.char1 ∧ str3 = str1.rstring ∧ str0 = str2

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
str1 = char0 + str4, # str0 = # str2, (char ' + str3) + str0 = str1 + str2, ∀ str0, str2, str1. # str0 = # str2 ∧ str3 + str0 = str1 + str2 → str3 = str1 ∧ str0 = str2 ⊦ str1 = "", char = char0 ∧ str3 = str4 ∧ str0 = str2

Rule argument

spec: string-data inst: name: elim seq: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0) subst: [str, char, str0] → [str1, char1, str] flags:

Additional information

Used simplifier rules: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0); (elim) Used terms: str1, char1, str