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Proof step 18 from proof ocs-put (
previous step
)
m0 < # x0, ∀ m, a1. m < # x0 → x0[m] = a1 ∨ #oc(a1, x0[m, a1]) = #oc(a1, x0) + 1 ⊦ #oc(a, x0[m0, a]) = 1 + #oc(a, x0), x0[m0] = a, a = a0, x0 = []
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: m0, a
Additional information
Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m0, a