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Proof step 4 from proof append-eqlen (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 5:
# 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

Rule argument

pos: left 3 terms: str1.rstring, str0, str2 abort: F discard: T computed: F

Additional information

Used terms: str1.rstring, str0, str2