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Proof step 3 from proof firstn-prefix-04 (previous step)

x = a + y ∧ x ≠ [], # x = (1 + # z) + n0 ⊦ a ' + firstn(n0, y) ⊑ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: prefix-rec-02: ⊦ b ' + x ⊑ b ' + y ↔ x ⊑ y; from specification list-last%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%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