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Proof step 13 from proof put-firstn-restn (previous step)

a0 ' + y0 + z = y + z0, # y = # y0, a0 ∈ y ∨ a0 ∈ z0 ⊦ z0[0, a] = a ' + z

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ x ≠ [] → x[0, a] = a ' + x.rest
Proof step 15:
z0 ≠ [], a0 ' + y0 + z = y + z0, # y = # y0, a0 ∈ y ∨ a0 ∈ z0 ⊦ a ' + z0.rest = a ' + z

Rule argument

spec: inst: name: put-zero-01 seq: ⊦ x ≠ [] → x[0, a] = a ' + x.rest subst: [a, x] → [a, z0] flags: rewrite, current

Additional information

Used simplifier rules: app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-empty: ⊦ ¬ a ∈ []; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: a, z0