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Proof step 4 from proof max-in (previous step)

i < max(i0 ' + x), i0 ≤ i, i0 < max(i0 ' + x), ∀ x0. # x0 < # (i + i0 + x) → (true → x0 = [] ∨ max(x0) ∈ x0) ⊦ max(i0 ' + x) ∈ x

User applied rule apply induction

And closed the goal!

Rule argument

[substlist-arg: [x0] → [i0 ' + x]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i < j → ¬ j < i], [CSIMPFORWARD : ⊦ i ≤ j ∧ j < j0 → i < j0], [CSIMPFORWARD : ⊦ i < j ∧ j ≤ j0 → i < j0], [CSIMPSEQ : ⊦ ¬ d ∈ x → (d ∈ d0 ' + x ↔ d = d0)], [CSIMPSEQ : ⊦ d ' + x ≠ []], [CSIMPSEQ : ⊦ # (x + y) = # x + # y], [CSIMPSEQ : ⊦ # d ' = 0 +1], [CSIMPSEQ : ⊦ (x + y) + z = x + y + z], [CSIMPSEQ : ⊦ d + x = d ' + x], [CSIMPSEQ : ⊦ m + n < m + n0 ↔ n < n0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (i0 ' + x)]]