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Proof step 9 from proof put-firstn-restn (previous step)

a0 ' + y0 + z = y + z0 ∧ # y = # y0 ∧ # y0 ≤ #(a0 ' + y0 + z) ⊦ (a0 ' + y0 + z)[# y0, a] = y + a ' + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# y = # y0, a0 ' + y0 + z = y + z0, a0 ∈ y ∨ a0 ∈ z0 ⊦ (y + z0)[# y0, a] = y + a ' + z

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list