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Proof step 2 from proof sublist-firsteq (previous step)

⊦ firstn(n, x) = a ' + y ↔ n = (n - 1) + 1 ∧ x.first = a ∧ firstn(n - 1, x.rest) = y

User applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] flags: []

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m