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Proof step 3 from proof firstn-append-01 (previous step)

∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → x0 = firstn(# x0 - # z, x0) + z, a0 + x0 = firstn(#(a0 + x0) - 1 + # z, a0 + x0) + a ' + z ⊦ a0 + x0 = firstn(#(a0 + x0) - # z, a0 + x0) + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a0 ' + x0 = firstn(# x0 - # z, a0 ' + x0) + a ' + z, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → x0 = firstn(# x0 - # z, x0) + z, a0 ∈ firstn(# x0 - # z, a0 ' + x0) ∨ a0 = a ∨ a0 ∈ z ⊦ a0 ' + x0 = firstn(# x0 + 1 - # z, a0 ' + x0) + z

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%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%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list