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Proof step 15 from proof to-vtbl-in (previous step)

∀ vols1, v. # vols1 < # vols0[v0, ebatbl] → true → (v ∈ to-vtbl(vols1) ↔ v ∈ vols1) ⊦ v ∈ to-vtbl(vols0) ↔ v ∈ vols0, v = v0, v0 ∈ vols0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [vols1, v] → [vols0, v]

Additional information

Used simplifier rules: ⊦ ¬ v ∈ vols → # vols[v, ebatbl] = # vols + 1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: vols0, v