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Proof step 93 from proof min-com (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 94:
⊦ x ≠ [] → min(a ' + x) = (a < min(x) ⊃ a;min(x))
Proof step 95:
min(x + a ' + x3) = (a < min(x3 + x) ⊃ a;min(x3 + x)), min(x3 + a0 ' + x) = min(x + x3), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ x + x3 = [], x3 + x = [], min(x + x3) = (a < min(x3 + x) ⊃ a;min(x3 + x)), a0 < min(x + x3), a0 < (a < min(x3 + x) ⊃ a;min(x3 + x)), a < min(x + x3)

Rule argument

spec: <toplevel> name: min-rec seq: ⊦ x ≠ [] → min(a ' + x) = (a < min(x) ⊃ a;min(x)) subst: [x, a] → [x3 + x, a] flags: <all paths>

Additional information

Used simplifier rules: min-base: ⊦ min(a ') = a;% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ [] + x = x; (s)% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x3 + x, a