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Proof step 17 from proof min-com (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 18:
a0 < min(x + a ' + x3), a < min(x3 + a0 ' + x), x + a ' + 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) ⊦ a = a0

Rule argument

subst: [x1, x2] → [x, a ' + x3]

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s)% ⊦ # a ' = 0 +1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s) Used terms: x, a ' + x3