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Proof step 5 from proof prefix-cut-l (conclusion)

⊦ p ≠ ε → (p + str0 ' ⊑ str1 ' + q ↔ (∃ p0, q1. p = str1 ' + p0 ∧ q = p0 + str0 ' + q1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
p + str0 ' ⊑ str1 ' + q ⊦ p = ε, ∃ p0, q1. p = str1 ' + p0 ∧ q = p0 + str0 ' + q1

Rule argument

Additional information

[Concretesimprules((Csimpseq( ⊦ (p + p0) + p1 = p + p0 + p1) Csimpseq( ⊦ str ' ⊑ str0 ' + p ↔ str = str0) Csimpseq( ⊦ p + p0 ⊑ p + p1 ↔ p0 ⊑ p1)))]