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Proof step 19 from proof put-firstn-restn (previous step)

y = a1 + y2, a0 ' + y0 + z = y + a ' + y1, # y = # y0, a0 ∈ y ∨ a0 = a ∨ a0 ∈ y1 ⊦ y1 = z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# y2 + 1 = # y0, y0 + z = y2 + a ' + y1 ⊦ y1 = z

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list