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Proof step 18 from proof size-consp (previous step)

∀ ps1, str. # ps1 < #(ps0 ++ p) → true → #(str + ps1) = # ps1 ⊦ #(str + ps0) = # ps0, p ∈ ps0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [ps1, str] → [ps0, str]

Additional information

Used simplifier rules: ⊦ ¬ p ∈ ps → #(ps ++ p) = # ps + 1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic[directory-tree-inst]%c: ⊦ m + n = n + m; from specification nat-basic[directory-tree-inst] Used terms: ps0, str