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Proof step 11 from proof recovery (previous step)

# avol0 = # lpt + m5, avol0[0] = mapped(buf), avol0[m2] = mapped(buf0), avol0[m0] = mapped(buf1), disk-layout(mksb(adr, m, m0, m2, m3, m4, m5), # lpt), ebm-vol-inv(subarray(avol0, m5, # lpt)), ebm-vol-inv(avol0), is-encoded(log0, buf1), is-encoded(lpt, lebs(avol0, m4, lptlebs(# lpt, LEB_SIZE))), is-encoded(mksb(adr, m, m0, m2, m3, m4, m5), buf), is-encoded-pad(ns0, m3, buf0), mapped-full?(avol0, m4, lptlebs(# lpt, LEB_SIZE)), m5 ≤ # avol0, avol0 ⊆ avol1, m1 ≠ LEB_SIZE → # buf1 = m1 ⊦ ∃ lpt1. m5 + # lpt1 = # avol0 ∧ mapped-full?(avol0, m4, lptlebs(# lpt1, LEB_SIZE)) ∧ is-encoded(lpt1, lebs(avol0, m4, lptlebs(# lpt1, LEB_SIZE)))

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: lpt

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic Used terms: lpt