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Proof step 5 from proof append-eqlen (previous step)

# str3 + 1 = # str1, (char ' + str3) + str0 = str1 + str2, # str3 = # str1.rstring ∧ str3 + str0 = str1.rstring + str2 → str3 = str1.rstring ∧ str0 = str2 ⊦ char ' + str3 = str1 ∧ str0 = str2

User applied rule cut formula

And obtained the following new goals

Proof step 6:
# str3 + 1 = # str1, (char ' + str3) + str0 = str1 + str2, # str3 = # str1.rstring ∧ str3 + str0 = str1.rstring + str2 → str3 = str1.rstring ∧ str0 = str2 ⊦ str1 = "", char ' + str3 = str1 ∧ str0 = str2

Rule argument

fma: str1 = ""

Additional information

Used simplifier rules: Length-empty: ⊦ # "" = 0;%len-zero: ⊦ # str = 0 ↔ str = "";%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 formulas: str1 = ""