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Proof step 3 from proof put-del1-in (previous step)

m ≤ # x0, ∀ m, a. x0[m] ≠ a ∧ m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ⊦ a ∈ (a0 ' + x0)[m, a] -1l a ↔ a = a0 ∨ a ∈ x0, (a0 ' + x0)[m] = a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.first ' + x.rest[n -1, a]
Proof step 5:
m ≠ 0 ∧ a0 ' + x0 ≠ [], m ≤ # x0, ∀ m, a. x0[m] ≠ a ∧ m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ⊦ a ∈ ((a0 ' + x0).first ' + (a0 ' + x0).rest[m -1, a]) -1l a ↔ a = a0 ∨ a ∈ x0, (a0 ' + x0)[m] = a

Rule argument

spec: inst: name: put-rec seq: ⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.first ' + x.rest[n -1, a] subst: [a, n, x] → [a, m, a0 ' + x0] flags: rewrite, current

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;%put-zero: ⊦ (b ' + x)[0, a] = a + x;%del1-y: ⊦ (a ' + x) -1l a = x;%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list Used terms: a, m, a0 ' + x0