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Proof step 52 from proof min-com (previous step)

min(x3 + a0 ' + x) = min(x + x3), min(x + a ' + x3) = min(x3 + x), a < min(x + x3), min(x3 + x) < a, min(x3 + x) < min(x + x3), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ a = min(x3 + x), x3 + x = [], x + x3 = [], a0 < min(x + x3), a0 < min(x3 + x)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [x1, x2] → [x3, x]

Additional information

Used simplifier rules: transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem%irreflexivity: ⊦ ¬ a < a; from specification oelem%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat Used terms: x3, x